$f (x, y)$ is given by $f (x, y) = (x^2 - 5x\cdot y)\cdot e^y$ These are the question to that function that I'm struggling with:


*

*Find the partial derivatives of first and second order of $f(x, y)$.

*Find the stationary points of $f(x, y)$ and determines for each point on 
it/they are a local maximum point, the local minimum point or saddle point.

*Is it possible to say something about the function has maximum and minimum values ​​based on the information you have found?


I've tried over and over and I'm getting real frustrated. It's a bonus problem that I really don't have to do, but I'd like to anyway. 
What I got on first problem:
First order: $f'_x(x,y) = (2x-5y)\cdot e^y$ and $f'_y(x,y)= -5x\cdot e^y$.
Correct?
 A: (1) It looks like your $f_x$ is correct (noting that one in general do not write $f_x'$ for the derivative. The subscript $x$ shows that you have taken the derivative with respect to $x$) However, your $f_y$ doesn't look quite right.
You have:
$$
f(x,y) = (x^2 - 5xy)e^y = x^2e^y - 5xye^y
$$
So
$$
f_y = x^2e^y-5xe^y - 5xye^y \quad\text{(product rule).}
$$
(2) To find the stationary points you need to solve the system of equations
$$
\begin{align}
f_x(x,y) = 0 \quad &\text{and}\quad f_y(x,y) = 0.\\
2x = 5y \quad&\text{and}\quad x^2-5x-5xy = 0\Rightarrow \\
x^2 - 5x - 5x(\frac{2}{5}x) &= 0 \Rightarrow \\
x(x -7) &= 0.
\end{align}
$$
You can probably solve this... 
(3) To classify the stationary points you compute the second order partial derivatives:
$$
f_{xx}, f_{yy}, f_{xy}, f_{yx}
$$
Then you compute the "discriminant":
$$
D = f_{xx}f_{yy} - f_{xy}f_{yx}.
$$
at the stationary points. Then you have 
$$
\begin{align}
D > 0 \text{ and } f_{xx} > 0 &\Rightarrow \text{local minimum} \\
D > 0 \text{ and } f_{xx} < 0 &\Rightarrow \text{local maximum} \\
D < 0 &\Rightarrow \text{saddle point}.
\end{align}
$$
If $D = 0$, you don't know.
