What are the extrema and saddle points of f(x,y) = 2x^(2) + (xy)^2 + 5x^2 - y/x?

Answer:

This function has no stationary points (are you sure that f(x,y)=2x^2+(xy)^2+5x^2−y/x is the one that you wanted to study?!).

Explanation:

According to the most diffused definition of saddle points (stationary points that are not extrema), you are searching for the stationary points of the function in its domain $D = \left\{\left(x , y\right) \in {\mathbb{R}}^{2} | x \ne 0\right\} = {\mathbb{R}}^{2} \setminus \left\{\left(0 , y\right) \in {\mathbb{R}}^{2}\right\}$.
We can now rewrite the expression given for $f$ in the following way: $f \left(x , y\right) = 7 {x}^{2} + {x}^{2} {y}^{2} - \frac{y}{x}$

The way to identify them is to search for the points that nullify the gradient of $f$, which is the vector of the partial derivatives:
$\nabla f = \left(\frac{\partial f}{\partial x} , \frac{\partial f}{\partial y}\right)$
Since the domain is an open set, we don't need to search for extrema eventually lying on the boundary, because an open sets contain no boundary points.

So let's compute the gradient of the function:
$\nabla f \left(x , y\right) = \left(14 x + 2 x {y}^{2} + \frac{y}{x} ^ 2 , 2 {x}^{2} y - \frac{1}{x}\right)$
This is null when the following equations are simultaneously satisfied:
$14 x + 2 x {y}^{2} + \frac{y}{x} ^ 2 = 0$
$2 {x}^{2} y = \frac{1}{x}$
We can turn the second into $y = \frac{1}{2 {x}^{3}}$ and substitute it into the first to get
$14 x + 2 x {\left(\frac{1}{2 {x}^{3}}\right)}^{2} + \frac{\frac{1}{2 {x}^{3}}}{x} ^ 2 = 0$
$14 x + \frac{1}{2 {x}^{5}} + \frac{1}{2 {x}^{5}} = 0$
$14 {x}^{6} + 1 = 0$

This can't be satisfied for $x \in \mathbb{R}$, so the gradient is never null on the domain. This means that the function has no stationary points!