Finding the maximum and minimum 
Can't understand how to find the maximum and minimum with the given definitions (with both x and y).. can someone explain step by step? 
 A: First of all compute the gradient of $f$:
$$\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial x}\right)=(1+y^2,2xy)$$
and solve $\nabla f=0$: this equation has no solutions. If it had you would need to pick the ones inside the domain you are considering and check their type by looking at the matrix of second derivatives.
Now you have to look at the boundary $x^2+y^2=4$, that you can parametrize taking $x=2\cos t$ and $y=2\sin t$: hence $f$ restricted to the boundary becomes 
$$f=2\cos t(1+4\sin^2 t)$$
$$\frac{df}{dt}=-2\sin t(1+4\sin^2 t)+2\cos t(8\sin t \cos t)=2\sin t(-1-4\sin^2 t+8\cos^2 t)$$
$$=\sin t(-1-4\sin^2 t+8-8\sin^2 t)=\sin t (7-12\sin^2 t)=\frac{1}{2}y(7-3y^2)=0 $$
From the last equation you can get the $y$ of the critical points and from $x^2+y^2=4$ get the $x$. In the end you obtain  6 points: $(\pm2,0)$, $(\pm\sqrt{5/3},\pm\sqrt{7/3})$. To check if they are maxima or minima you have just to look at the sign of the second derivative (in $t$) of $f$ restricted to the boundary.

