This week's problem was correctly answered by chisigma, lfdahl, magneto, MarkFL, and Pranav. You can find Mark's solution below.
[sp]First, we need to determine where $y^2>x^2$ in the square region over which we are integrating. We find that the diagonal of the square along the line $y=x$ divides the region into two areas. The right isosceles triangle with vertices $(0,0),\,(0,1),\,(1,1)$ is where $y^2\ge x$ and the right isosceles triangle with vertices $(0,0),\,(1,1),\,(1,0)$ is where $x^2\ge y^2$.
We will treat the triangular region where $x^2\ge y^2$ as a Type I region and the region where $y^2\ge x$ as a Type II region.
Thus, we may state:
$$I=\int_0^1\int_0^1 \exp\left(\max\left(x^2,y^2 \right) \right)\,dy\,dx=\int_0^1\int_0^x e^{x^2}\,dy\,dx+\int_0^1\int_0^y e^{y^2}\,dx\,dy$$
Rewriting the iterated integrals, we have:
$$I=\int_0^1 e^{x^2}\left[\int_0^x \,dy \right]\,dx+\int_0^1e^{y^2}\left[\int_0^y \,dx \right]\,dy$$
Applying the FTOC on the inner integrals, we obtain:
$$I=\int_0^1 xe^{x^2}\,dx+\int_0^1 ye^{y^2}\,dy$$
Observing that if we change the dummy variables of integration both to $u$, we find:
$$I=2\int_0^1 ue^{u^2}\,du$$
Using the substitution:
$$v=u^2\,\therefore\,dv=2u\,du$$
We find:
$$I=\int_0^1 e^{v}\,dv=\left.e^v \right|_0^1=e-1$$
Hence, we may conclude that:
$$I=\int_0^1\int_0^1 \exp\left(\max\left(x^2,y^2 \right) \right)\,dy\,dx=e-1$$[/sp]