Let $f:\mathbb{R^2}\rightarrow \mathbb{R}, f(x,y)=(x+y)\exp{(x^2+y^2)}, \ x^2+y^2 \leq1$. Find the maximum and minimum of the function $f.$
I computed the partial derivates of f as: $$\partial_xf=\exp(x^2+y^2)(2x^2+2xy+1),\ \partial_yf=\exp(x^2+y^2)(2y^2+2xy+1) \\ \partial_{xx}f=2\exp(x^2+y^2)(2x^3+2x^2y+3x+y), \ \partial_{yy}=2\exp(x^2+y^2)(2y^3+2y^2x+3y+x) $$ I assumed that the exponential function behaves exactly the same for multivariable function to end up with $\exp(x^2+y^2)>0$ alongside the polynomials of the form $2x^2+2xy+1$. If I'm not mistaken the polynomials describe a hyperbola and the only solution of the polynomials are: $\{x=\sqrt2,y=-\frac12\}, \{x=-\sqrt2,y=\frac12\}$ which do not satisfy the given condition in the question so I came to the conclusion that function $f$ has no extreme values. However I'm not entirely sure of my answer and would appreciate if someone could at least verify it for me.
Since the function $f$ is continuous over the compact set $x^2+y^2\leq 1$ then the extreme points exist! Your work shows that there are no stationary points (the two hyperbolas do not intersect!) which implies that the extreme points have to be located along the boundary $x^2+y^2=1$.
Note that in polar coordinates $(r,\theta)$ the function can be written as $$f(x,y)=r(\cos(\theta)+\sin(\theta))\exp(r^2)=\sqrt{2}r\exp(r^2)\,\sin(\theta+ \pi/4).$$ Therefore along each circle $x^2+y^2=r$ the function attains its maximum value $\sqrt{2}r\exp(r^2)$ for $\theta=\pi/4$ and its minimum value $-\sqrt{2}r\exp(r^2)$ for $\theta=5\pi/4$. Since $r\exp(r^2)$ is strictly increasing in $r\in [0,1]$, we find that the maximum/minimum value of $f$ over the disc $x^2+y^2\leq 1$ is attained for $r=1$.
Therefore we may conclude that the maximum value is $f(1/\sqrt{2},1/\sqrt{2})=\sqrt{2}e$ and the minimum value is $f(-1/\sqrt{2},-1/\sqrt{2})=-\sqrt{2}e$.