I have this:
Let $X\sim U(0,1)$, $Y\sim U(X,1)$. What is the distribution of variable $Y$?
My answer: I use a geometric approach since everything happens in the square $(0,1)\times (0,1)$, see the image.
so I found that
$$F(y)= \mathbb{P}\{ Y \leq y \}= \begin{cases} 0 & y<0 \\ y^2 & 0 \leq y\leq 1\\ 1 & y>1 \end{cases} $$ then $$f(y)= \begin{cases}2y & 0\leq y\leq 1\\ 0 & \text{otherwise.} \end{cases} $$ Is this approach correct?
A sanity check: verify you get the right expectation. (This is not sufficient, but at least it's a necessary condition for correctness).
Since $Y\sim U(X,1)$, we have $$\mathbb{E}[Y\mid X] = \frac{1}{2}(1+X)$$ and therefore $$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y\mid X]] = \frac{1}{2}(1+\mathbb{E}[X]) = \frac{1}{2}(1+\frac{1}{2}) = \frac{3}{4}$$ since $X\sim U(0,1)$ (for the second-to-last equality).
Your solution, however, yields $$\mathbb{E}[Y] = \int_{-\infty}^\infty dy f(y)y=\int_{0}^1 2y^2dy = \frac{2y^3}{3}\Big|_0^1=\frac{2}{3} $$ so your answer cannot be correct.
A derivation:
Fix any $t\in\mathbb{R}$. First of all, it is obvious that $F_Y(t) = 0$ if $t\leq 0$ and $F_Y(t) = 1$ if $t\geq 1$, so we can focus on $t\in(0,1)$. Then, we have $$ F_Y(t) = \int_0^1 dx \int_x^t dy \frac{1}{1-x}\mathbb{1}_{\{t\geq x\}} = \int_0^t dx \int_x^t dy \frac{1}{1-x} = \int_0^t dx \frac{t-x}{1-x} $$ and after computation, we get $$ F_Y(t) = \begin{cases} 0 &t\leq 0\\ t+(1-t)\ln(1-t) & t\in(0,1)\\ 1 & t\geq 1 \end{cases} $$ (which after verification is consistent: $F_Y(0^+)=0, F_Y(1^-)=1$, and $F_Y$ is non-decreasing and continuous.)
Deriving this CDF, we obtain $$ f_Y(t) = -\ln(1-t) \mathbb{1}_{(0,1)}(t),\qquad t\in\mathbb{R} $$ which passes the (necessary) test $$ \int_{-\infty}^\infty t f_Y(t) dt = -\int_0^1 t\ln(1-t) dt = \frac{3}{4}. $$
Plot of the results obtained in the derivation:
PDF $f_Y$ of $Y$
CDF $F_Y$ of $Y$