Finding the probability of $Y>1/4$ when conditioned on $X=x$.

Just use Bayes' Rule, and the Law of Total Probability

$$begin{align}mathsf P(Y>amid X=x)&=int_a^infty f_{small Ymid X}(ymid x)~mathrm d y\[2ex]&=int_a^infty dfrac{f_{small X,Y}(x,y)}{f_{small X}(x)}~mathrm d y\[2ex] &=dfrac{int_a^infty f_{small X,Y}(x,y)~mathrm d y}{int_Bbb R f_{small X,Y}(x,y)~mathrm d y}\[4ex]text{so since}~f_{small X,Y}(x,y)&=2xmathrm e^{x^2}mathrm e^{-y}mathbf 1_{0lt xlt 1}mathbf 1_{x^2lt y}\[4ex]mathsf P(Y>1/4mid X=x)&=require{cancel}dfrac{cancel{2xmathrm e^{x^2}}}{cancel{2xmathrm e^{x^2}}}cdotdfrac{int_{max{1/4,x^2}}^infty mathrm e^{-y}~mathrm d y}{int_{x^2}^infty mathrm e^{-y}~mathrm d y}cdotmathbf 1_{0<x<1}\[2ex]&=dfrac{int_{1/4}^inftymathrm e^{-y}~mathrm d y}{int_{x^2}^infty mathrm e^{-y}~mathrm d y }mathbf 1_{0<x<1/2}+mathbf 1_{1/2leqslant xle 1}\[2ex]&~~vdotsend{align}$$

The way the conditional distribution of $Y$ given $X$ is found, is to integrate fixing the $x$ in $f$, so we are within the region $X = x$. However, note that if $P$ has marginal density zero at $x$, then the conditional density will also be zero.

That is, $$ P(Y in A| X = x) = frac{int_A f(x,y)dy}{int_{mathbb R} f(x,y)dy} $$

for any subset $A subset R$ and $x$ such that the denominator is non-zero. Otherwise the conditional probability is zero. In our case, $A = [frac 14 , infty)$.

Now, $int_{mathbb R} f(x,y)dy = 0$ for $x leq 0$ or $x geq 1$, so then the expression is zero. Else, $$int_{mathbb R} f(x,y) dy = int_{x^2}^infty 2xe^{x^2}e^{-y}dy = 2xe^{x^2}int_{x^2}^infty e^{-y}dy$$ is easily calculated.

Then, in similar fashion, $$ int_{frac 14}^infty f(x,y)dy = 2xe^{x^2}int_{max{x^2,frac 14}}^infty e^{-y}dy $$

which is again very easily calculated, leading to the answer, following the cancellation of $2xe^{e^{x^2}}$ when we are dividing top by bottom.