A figure indicating the range of $(X, Y)$ often helps to determine the integral bounds and solve problems like this.
In this problem, the range of $(X, Y)$ is given by $0 \leq x^2 + y^2 \leq 1$, which is the unit circle (showing below).
Since the area of the unit circle is $\pi \times 1^2 = \pi$ and given that
$$f(x,y) = \frac{1}{\pi} \quad \text{if } x^2 + y^2 \leq 1,$$
and 0 otherwise, we conclude that $(X, Y)$ is uniformly distributed over the unit circle (think about why this is true).
Next, we want to find $P(Y > \left|X\right|)$. Since $(X, Y)$ is uniformly distributed, the problem reduces to finding the ratio between areas of the region defined by the event $\{Y > \left|X\right|\}$ (i.e. highlighted in light blue) and the unit circle. From the figure we can see that the highlighted region is in fact a quarter of the unit circle (the angle between $y = x$ and $y = -x$ is 90 degrees), we have
$$P(Y > \left|X\right|) = \frac{\frac{1}{4}\pi}{\pi} = \frac{1}{4}.$$
Remark: This approach makes use of the property of uniform distribution, which bypasses calculating the integral. If we do want to solve this problem using the integral approach, note that
\begin{align*}
P(Y > \left|X\right|) &= \iint_{\{y > \left|x\right|\}} f(x,y) dxdy\\
&= \iint_{\{y > \left|x\right|, \ x^2 + y^2 \leq 1\}} \frac{1}{\pi} dxdy\\
&= \frac{1}{\pi} \iint_{\{y > \left|x\right|, \ x^2 + y^2 \leq 1\}} dxdy\\
&= \frac{1}{\pi} \int_{\pi/4}^{3\pi/4} \int_0^1 rdrd\theta \quad (\text{polar coordinate})\\
&= \frac{1}{\pi} \int_{\pi/4}^{3\pi/4} \left[\frac{1}{2}r^2\right]\bigg|_0^1 d\theta\\
&= \frac{1}{\pi} \int_{\pi/4}^{3\pi/4} \frac{1}{2} d\theta\\
&= \frac{1}{2\pi} \left[\frac{3\pi}{4} - \frac{\pi}{4}\right]\\
&= \frac{1}{2\pi} \cdot \frac{\pi}{2}\\
&= \frac{1}{4}.
\end{align*}
More details on polar coordinate transformation can be found here.
Best Answer
With these types of problems it is always helpful to draw pictures. First, we have for the joint density of $X$ and $Y$: $$ f_{XY}(x,y)=e^{-x}e^{-y},\quad 0<x,y<\infty. $$ Now draw the region $0<y<x<2$:
The probability in question can now easily be written as an integral in the form $$ \mathsf P(Y<X<2)=\int_0^2\int_0^x e^{-x}e^{-y}\,\mathrm dy\mathrm dx=\frac{\left(e^2-1\right)^2}{2 e^4}\approx 0.75. $$