Using Lagrange multipliers, find the extreme values of $U(x,y)=e^{x+y}$ on the surface $x^2+xy+y^2=1$
$\nabla U(x,y)=\lambda\nabla f(x,y) $ where $f(x,y)=x^2+xy+y^2-1$ (the constraint surface).
$$(e^{x+y},e^{x+y})=\lambda(2x+y,2y+x)$$ Then $$e^{x+y}=\lambda(2x+y)$$ $$e^{x+y}=\lambda(2y+x)$$
Therefore: $\lambda >0$, $\lambda(2x+y)=\lambda(2y+x) \Rightarrow 2x+y=2y+x \Rightarrow x=y$
If $x=y$, then substituting in our constraint equation: $x^2+x(x)+y^2=1 \Rightarrow 3x^2=1 \Rightarrow x=\pm\frac{1}{\sqrt{3}}$ Since, $x=y$, our extrema are $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$ and $(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}})$
You immediately get $\lambda>0$ and $2x+y=2y+x$, so $x=y$. Plugging this into the constraint, you get $3x^2=1$, so $x=\pm 1/\sqrt{3}$.