Answer
Maximum:$f(1,1)=e$, Minimum: $f(-1,1)=-e$
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y)=\lambda \nabla g(x,y)$
This yields $\nabla f=\lt e^y,xe^y \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y \gt$
Using the constraint condition $x^2+y^2=2$ we get, $e^y=\lambda 2x, xe^y=\lambda 2y$
After solving, we get $x=\pm 1$
Since, $g(x,y)=x^2+y^2=2$ gives $y=-2,1$
When, $y=-2$, thus $x^2+y^2=2 \implies x^2=-2$; which is incorrect point.
Hence, Maximum:$f(1,1)=e$, Minimum: $f(-1,1)=-e$
