Answer:
Posterior of $\sigma^2|Y_1,..., Y_n$ is an instance of inverse gamma distribution with the probability density
$$ p(\sigma^2|Y_1,...,Y_n) = \frac{\beta^\alpha}{\Gamma(\alpha)} (\sigma^2)^{-\alpha+1}\exp(-\frac{\beta}{\sigma^2}), $$
where
\begin{align}
\alpha:=\frac{\nu_0+n}{2}, \quad& \beta:=\frac{\nu_0\sigma_0^2+n(\hat \sigma^2 + \frac{k_0}{k_0+n}(\bar Y - \mu_0)^2)}{2},\\
\bar Y := \frac{1}{n}\sum_i Y_i, \quad &\hat \sigma^2 := \frac{1}{n}\sum_i(Y_i-\bar Y)^2.
\end{align}
The posterior of $\mu|Y_1,...,Y_n$ is a shifted and scaled Student's t-distribution. It's probability density function can be written down as
$$ p(\mu|Y_1,...,Y_n) = \frac{\Gamma \left(\frac{\nu+1}{2} \right)} {\sqrt{\nu\pi\gamma}\,\Gamma \left(\frac{\nu}{2} \right)} \left(1+\frac{(\mu-\mu')^2}{\nu \gamma} \right)^{-\frac{\nu+1}{2}},$$
where $\Gamma(x)$ is the Gamma function, $\nu:=\nu_0+n$ is the number of degrees of freedom, $\gamma:=\frac{\beta}{\alpha(n+k_0)}$ is the scale parameter $\mu' := \frac{n}{n+k_0}\bar Y + \frac{k_0}{n+k_0}\mu_0 $ is the mean of the interim posterior $\mu|\tau,\mathbf{Y}$:
\begin{equation}
\mu|\tau,\mathbf{Y} \sim N(\mu', ((n+k_0)\tau)^{-1}). \qquad\qquad (\star)
\end{equation}
Detailed steps:
(There is a great variety of sources online, one of which I am reproducing almost without changes: lectures of Michael I. Jordan from Berkeley.)
Derivation is quite straightforward, once we introduce notation $\tau := \frac{1}{\sigma^2}$, $\alpha_0 := \frac{\nu_0}{2}$, $\beta_0 := \alpha_0 \sigma^2_0$, and recognize in the stated prior the following hierarchical model:
\begin{align}
Y_i &\sim N(\mu, \tau^{-1})\\
\mu &\sim N(\mu_0, (k_0\tau)^{-1})\\
\tau &\sim Gamma(\alpha_0,\beta_0)\\
\end{align}
where $Gamma$ stands for Gamma distribution with the probability density function
$$p(\tau|\alpha_0,\beta_0) = \frac{\beta_0^{\alpha_0}}{\Gamma(\alpha_0)} \tau^{\alpha_0-1}\exp(-\tau \beta_0).$$
We are looking for the posterior $\mu| \mathbf{Y}$ and $\tau | \mathbf{Y}$, where $\mathbf{Y}:= (Y_1,...,Y_n) $.
Obtaining posterior $\mu|\mathbf{Y}$ is a matter of taking an expectation of pdf of the interim posterior $(\star)$ with respect to the posterior $\tau|\mathbf{Y}$:
\begin{equation}
p(\mu|\mathbf{Y}) = \int_0^\infty p(\mu|\tau, \mathbf{Y}) p(\tau|\mathbf{Y})d \tau \quad (*)
\end{equation}
So the first step would be to obtain the posterior $\tau| \mathbf{Y}$, which is just the marginal of the joint posterior $\mu,\tau |\mathbf{Y}$:
\begin{align}
p(\tau,\mu|\mathbf{Y})
\propto & \prod_i p(Y_i|\mu,\tau)\cdot p(\mu|\tau)\cdot p(\tau) \\
\propto & \tau^{n/2} \exp\left(-\frac{\tau}{2}\sum_i(Y_i-\mu + \bar Y - \bar Y)^2\right) \cdot \tau^{1/2} \exp\left( -\frac{k_0 \tau}{2}(\mu-\mu_0)^2 \right) \cdot \tau^{\alpha_0-1} \exp(-\beta_0\tau) \\
\propto & \tau^{\alpha_0 + \frac{n}{2}-1}\exp\left(-\tau(\beta_0 + \frac{1}{2}\sum(Y_i-\bar Y)^2) \right) \tau^{1/2} \cdot \exp\left(-\frac{\tau}{2}(k_0(\mu-\mu_0)^2+n(\bar Y - \mu)^2)\right)
\end{align}
In the last expression we can factorize a kernel of a normal out of the second term (on the right of $\cdot$):
\begin{align}
&\exp\left(-\frac{\tau}{2}(k_0(\mu-\mu_0)^2+n(\bar Y - \mu)^2)\right) =\\
&= \exp\left(-\frac{\tau}{2}((k_0+n)\mu^2-2(k_0\mu_0 + n\bar Y) \mu + k_0 \mu_0^2+n\bar Y^2)\right)\\
&= \exp\left(-\frac{\tau}{2}((k_0+n)(\mu^2-2\frac{k_0\mu_0 + n\bar Y}{k_0+n}\mu + {\mu'}^2) - (k_0+n){\mu'}^2 + k_0 \mu_0^2+n\bar Y^2)\right)\\
&= \exp(-\frac{\tau}{2}(k_0+n)(\mu - \mu')^2) \cdot \exp\left(\frac{\tau}{2}( \frac{k_0^2 \mu_0^2 + 2 n k_0 \mu_0 \bar Y + n^2 \bar Y^2}{n+k_0} - k_0 \mu_0^2 -n \bar Y^2)\right)\\
&= \tau^{1/2} \exp(-\frac{\tau}{2}(k_0+n)(\mu - \mu')^2) \cdot \tau^{-1/2}\exp\left(-\frac{n k_0 \tau}{2(n+k_0)} (\bar Y -\mu_0)^2\right)
\end{align}
The first term in the above product is going to integrate to $\sqrt{\frac{2\pi}{k_0+n}}$ (pdf of $N(\mu', \frac{1}{(n+k_0)\tau})$) and may be neglected, whereas the second term will be factorized leaving us with the following posterior for $\tau| \mathbf{Y}$:
\begin{equation}
p(\tau|\mathbf{Y}) \propto \tau^{\alpha_0 + \frac{n}{2}-1} \exp(-\tau \left(\beta_0 + \frac{1}{2}\sum_i(Y_i-\bar Y)^2 + \frac{n k_0}{2(n+k_0)}(\bar Y -\mu_0)^2\right))
\end{equation}
in which the kernel of a Gamma distribution is easily recognizable, i.e.
$$\tau|\mathbf{Y} \sim Gamma(\alpha,\beta)$$
where $\alpha := \alpha_0 + \frac{n}{2}$ and $\beta := \beta_0 + \frac{1}{2}\sum_i(Y_i-\bar Y)^2 + \frac{n k_0}{2(n+k_0)}(\bar Y -\mu_0)^2$.
Finally compute the expectation $(*)$:
\begin{align}
p(\mu|\mathbf{Y}) = & \int_0^\infty \frac{{\beta}^{\alpha}}{\Gamma(\alpha)}\tau^{\alpha-1} \exp(-\tau \beta) \cdot \frac{(n+k_0)^{1/2}\tau^{1/2}}{\sqrt{2 \pi}} \exp\left(-\frac{n+k_0}{2}\tau (\mu - \mu')^2 \right) d\tau \\
\propto & \int_0^\infty \tau^{\alpha+\frac{1}{2}-1} \exp\left(-\tau \beta - \tau \frac{n+k_0}{2}(\mu - \mu')^2 \right) d\tau \quad (**)\\
\propto & \Gamma(\alpha + \frac{1}{2}) \left(\beta + \frac{n+k_0}{2}(\mu-\mu')^2\right)^{-\alpha-\frac{1}{2}} \\
\propto & (1 + \frac{1}{2\alpha}\frac{(\mu-\mu')^2}{\frac{\beta}{(n+k_0)\alpha}})^{-\frac{2\alpha+1}{2}}
\end{align}
In the integrand in expression $(**)$ we see the kernel of $Gamma(\alpha+\frac{1}{2}, \beta + \frac{n+k_0}{2}(\mu - \mu')^2)$ which integrates to the expression in which one can easily recognize the kernel of a Student's t-distribution with mean $\mu'$, scale parameter $\frac{\beta}{(n+k_0)\alpha}$ and $2\alpha$ degrees of freedom.
