I know that variables $X$ and $Y$ are independent if $f(y|X=x)=f_Y$. In this case the joint pmf is given as $x +\frac{3y^2}{2}, \text{whenever } 0 \le x,y \le 1$ So: $$f_Y=\int^1_0 x +\frac{3y^2}{2} dx = \left. \frac{x^2}{2}+\frac{3xy^2}{2} \right|^1_0 = \frac{1+3y^2}{2}$$
$$f_X=\int^1_0x+\frac{3y^2}{2}dy=\left. xy+\frac{y^3}{2} \right|^1_0=x+\frac{1}{2}=\frac{2x+1}{2}$$
Then $$f(y|X=x)= \left( x+\frac{3y^2}{2} \right) :\left(x+\frac{1}{2}\right)$$ So does $$\frac{1+3y^2}{2}\stackrel{?}{=}\left(x+\frac{3y^2}{2}\right) : \left( x+\frac{1}{2}\right) $$
$$x+3xy^2+\frac{1}{2}+\frac{3y^2}{2}\stackrel{?}{= } 2x+3y^2$$
$$2x+6xy^2+1+3y^2\stackrel{?}{=}4x+6y^2$$
or $$12xy^2+2 \stackrel{?}{=}2x$$
It looks like no, but is impossible for me to tell. And I bet there is an easer way to tell whether these variables are independent. What could be the way to solve this?
You are given joint PDF $f_{X,Y}(x,y)$ and computed marginal PDF's $f_X(x)$ and $f_Y(y)$. By definition, $X$ and $Y$ are independent if joint PDF is a product of marginal PDF's, $$f_{X,Y}(x,y)=f_X(y)f_Y(y)$$