$f_X(x) = \frac{1}{2}e^{\frac{-x}{2}}$ and $f_{Y|X}(y|x) = I_{[0;x^2]}$ (Uniform continuous from $0$ to $x^2$).
I tried finding the joint distribution by using $f(X,Y) = f(Y|X) * f(X)$ and then integrating that from $x$ to find $f(Y)$ but the answer I got was an incomplete gamma function, which makes me believe I made a mistake somewhere in there. Any help or tips is appreciated.
The joint pdf of $X$ and $Y$ is $$f(x,y)=\frac1{2x^2}e^{-\frac x2}$$ if $0\leq y \leq x^2$ and $0$ otherwise. That is, $f(x,y)=0$ if
$$x\leq \sqrt y.$$ As a result, we have to integrate the joint distribution with respect to $x$ from $\sqrt y$ to $\infty$:
$$f_Y(y)=\int_{\sqrt y}^{\infty}\frac1{2x^2}e^{-\frac x2}\ dx.$$
With the help of Wolfram Alpha, you can get something containing the $\Gamma$ function:
$$f_Y(y)=\frac{e^{-\sqrt{y}/2}}{2\sqrt y}-\frac{\Gamma(0,\sqrt y/2)}{4}.$$
However ugly this is, it is just a nice pdf whose integral from $0$ to $\infty$ is $1$.