I'm a bit confused about how to calculate conditional distributions. I know that:
$f_{Y|X=x}\left(x,y\right)=\frac{f_{X,Y}\left(x,y\right)}{f_X\left(x\right)}$
Does this mean that for the following joint distribution:
$f(x,y)=\left\{\begin{matrix}\frac{3}{2}&x\in\left(0,1\right)\mathrm{\ and\ } x^2\le y\le1\\0&else\\\end{matrix}\right.$
the calculation to find $f_{Y|X=x}(x,y)$ is:
$f_X\left(x\right)=\int_{\mathbb{R}} f\left(x,y\right)dy=\int_{0}^{1}{\frac{3}{2}dy}=\frac{3}{2}$
combined with:
$f_{Y|X=x}\left(x,y\right)=\frac{f_{X,Y}\left(x,y\right)}{f_X\left(x\right)}$
leading to:
$f_{Y|X=x}\left(x,y\right)=\frac{\frac{3}{2}}{\frac{3}{2}}=1$
If I'm wrong I'd appreciate some help with the general guidelines for finding marginal distributions
Your calculation of $f_X$ is wrong . $f_X(x)=\int_{x^{2}}^{1} \frac 3 2 \, dx =\frac 3 2 (1-x^{2})$.
When you calculate $f_X(x)$ you fix $x$ and integrate $f_{X,Y}$ w.r.t. $y$. You are integrating $\frac 3 2$ all the way from $0$ to $1$ but remember that when $y <x^{2}$ the function $f_{X,Y}(x,y)$ is given to be $0$. Hence you get integral of $\frac 3 2$ from $x^{2}$ to $1$.