There are some specific cases where it is true, such as a bivariate normal, then $f(Y)$ and $f(Y|X)$ are both normal.
But consider the case where $f(Y|X)$ is normal (with mean depending on $X$) and $X$ follows a uniform distribution. Then $f(Y)$ is not normally distributed.
There are also distributions where the marginals ($f(X)$ and $f(Y)$) are both $\text{uniform}(0,1)$, but there is a hole in the square where the probability is $0$, so the conditional $f(Y|X)$ would not be uniform and for some values of $X$ would be disjoint.