I solved the following exercise and want to know, if I did it correct
Consider a 2-dimensional random vector $(X,Y)$ with density $$f_{X,Y}(x,y)=c\frac{y}{x^3}1_{\{0<x\leq 1\}}1_{\{0<y\leq x^2\}}$$ a) Compute $c$, $f_X$ of $X$, $P[X\leq 1/2]$
b) Compute $E[X/Y]$, $E[Y\mid X=1/2]$
c) Are $X$ and $Y$ independent?
a) use $z=\sqrt{y}$ and $2\sqrt{y}\mathrm{d}z=\mathrm{d}y$ to get $$2c\int_0^1 \int_0^x \frac{z^3}{x^3}\mathrm{d}z\mathrm{d}x=\frac{c}{2}\int_0^1 x\mathrm{d}x=\frac{c}{4}\implies c=4$$ $$f_X=2x1_{\{0<x\leq 1\}}$$ $$P[X\leq 1/2]=2\int_0^{1/2} x\mathrm{d}x=2[\frac{x^2}{2}]^{1/2}_0=\frac{1}{4}$$ b) $$E[X/Y]=\int \int \frac{x}{y} f_{X,Y}(x,y) \mathrm{d}y\mathrm{d}x=4\int_0^1 \int_0^{x^2} \frac{x}{y}\frac{y}{x^3}\mathrm{d}y\mathrm{d}x$$ $$=4\int_0^1 \int_0^{x^2} \frac{1}{x^2}\mathrm{d}y\mathrm{d}x = 4\int_0^1 \frac{x^2}{x^2}\mathrm{d}x =4$$ $$E[Y\mid X=1/2]=\int y \frac{f_{X,Y}(1/2,y)}{f_{X}(1/2)}\mathrm{d}x \mathrm{d}y=\int y \frac{f_{X,Y}(1/2,y)}{f_{X}(1/2)}\mathrm{d}x \mathrm{d}y = 4\int_0^{1/4} y\cdot \frac{y}{1/8} \cdot (\frac{2}{2})^{-1}\mathrm{d}y$$ $$= 32 \int_0^{1/4} y^2\mathrm{d}y = \frac{32}{192}$$
c) I would say no because of the indicator function of $y$ depending on $x^2$, but I am not sure how to prove this. $$f_Y=\frac{f_{X,Y}}{f_{X}}=\frac{4\frac{y}{x^3}1_{\{0<x\leq 1\}}1_{\{0<y\leq x^2\}}}{2x1_{\{0<x\leq 1\}}}=2\frac{y}{x^4}1_{\{0<y\leq x^2\}}$$
Since the joint pdf factorises as $$f_{X,Y}(x,y)=\frac{2y}{x^4}\,\mathbf1_{0<y<x^2}\underbrace{2x\mathbf1_{0<x<1}}_{f_X(x)}\quad,$$
this automatically shows that the conditional density of $Y\mid X$ is
$$f_{Y\mid X=x}(y)=\frac{2y}{x^4}\mathbf1_{0<y<x^2}$$
This distribution depends on $X$, so that rules out independence of $X$ and $Y$.
And from the pdf you get $$E(Y\mid X=x)=\int_0^{x^2}\frac{2y^2}{x^4}\,dy=\frac{2x^2}{3}$$
This matches with your answer. The rest also looks correct.