Local minimum and maximum of two variable implicit function What is local minimum and maximum of implicit function $F(x,y)=x^4+y^4−x^2 −y^2=0$?
I calculated first and second derivative:
$$\frac{\partial F(x, y(x))}{\partial x}=4x^3+4y^3y'-2x-2yy'$$
$$\frac{\partial F^2(x, y(x))}{\partial x^2}=12x^2+12y^2(y')^2+12y^3y''-2(y')^2-2yy''-2$$
$$y'=\frac{2x-4x^3}{4y^3-2y}$$
$$y''=\frac{(y')^2-6x^2-6y^2(y')^2+1}{6y^3-y}$$
Then I don't know how to continue. How to find stationary points? How to decide if stationary point is local minimum or maximum?
 A: When differentiating implicitly you should get that:
\begin{align*}
 y'(x) = - \frac{\frac{\partial F}{\partial x}(x,y(x))}{\frac{\partial F}{\partial y}(x,y(x))} = \frac{2x-4x^3}{2y-4y^3}
\end{align*}
Now at an extremum we have that $y'(x)=0$. Lets assume that $2y-4y^3 \neq 0 $ (if we want $y$ to be an implicit function of $x$ then this is a demand). Then $y'(x)=0 \Leftrightarrow 2x-4x^3 =0 \Leftrightarrow 2x(1-2x^2)=0 \Leftrightarrow x =0 \vee x= \pm \sqrt{\frac{1}{2}}$.
Now just check the values of $y$ at these $x$-values (and possible endpoints of the domain of definition of $y$).
A: Given the function $f(x) = x^4+y^4-x^2-y^2$, you find the stationary points by first taking the partial derivatives:
$$
f_x' = 4x^3 - 2x\\
f_y' = 4y^3-2y
$$
The stationary points are those in which all partial derivatives evaluate to 0. That means you solve the following system of equations:
$$
\begin{cases}
0 = 4x^3 - 2x\\
0 = 4y^3 - 2y
\end{cases}
$$
It's a non-linear system in two variables so there is no standard solution method. Instead, you must begin by factoring:
$$
\begin{cases}
0 = 4x^3 - 2x\\
0 = 4y^3 - 2y
\end{cases}
\iff
\begin{cases}
0 = 2x(2x^2 - 1)\\
0 = 2y(2y^2 - 1)
\end{cases}
$$
Note that both equations have the same structure. For the first equation, it is obvious that $x = 0$ satisfies it. And if $x \neq 0$, then $2x^2 - 1$ must be 0 for it to be satisfied. So you solve that:
$$
2x^2 - 1 = 0 \iff 2x^2 = 1 \iff x^2 = \frac{1}{2} \iff x = \pm\frac{1}{\sqrt{2}}
$$
So for the first equation you have three solutions: $x_1 = 0, x_2 = -a, x_2 = a$, where $a = \frac{1}{\sqrt{2}}$ (this variable assignment reduces typing). Since the second equation is the same, it has the same three solutions: $y_1 = 0, y_2 = -a, y_3 = a$. 
Now to find all solutions to the equation system, just combine these roots: $(0, 0), (0, a), (0, -a), (a, 0), (a, a), (a, -a), (-a, 0), (-a, a)$ and $(-a, -a)$.
To find out whether these points are minimums or maximums, calculate the second order derivatives of the function:
$$
f_{xx}' = 12x^2 - 2\\
f_{yy}' = 12y^2 - 2\\
f_{xy}' = 0
$$
Then use the "second derivate test." For each point, you calculate like the following:
$$
D(0,0) = f_{xx}'(0, 0)f_{yy}'(0,0)-f_{xy}(0,0)^2 = (-2)(-2) - 0^2 = 4
$$
Because $D(0,0)>0$ and $f_{xx}'(0,0) = -2 < 0$ it is shown that $(0,0)$ is a local maximum. If $p$ is your point, then the rules are
$$
D(p) > 0 \land f_{xx}'(p) > 0 \implies \text{minimum}\\
D(p) > 0 \land f_{xx}'(p) < 0 \implies \text{maximum}\\
D(p) < 0 \implies \text{saddle point}\\
D(p) = 0 \implies \text{annoying special case}
$$
You can calculate the other points in exactly the same way.
