Find the maximum values of the function $f(x,y,z)=x^2y^2z^2$ subject to the constraint $x^2+y^2+z^2=289$. I have already figured out the majority of the problem. I took the derivative of $x$ and got: $x=0,y=0,z=0$. Initially I put that there were infinitely many solutions, but that is not true. I do not know where I went wrong as I took the derivatives in respect to $x$ and subtracted from the derivative of x to the $g(x)$. However, it gave me zero for all of my solutions.
I successfully found the minimum though.
 A: With AM-GM$$x^2+y^2+z^2=289\ge3\cdot \sqrt[3]{x^2y^2z^2}$$ Thus $$\bigg(\frac{289}{3}\bigg)^3\geq x^2y^2z^2$$
Equality holds iff $\pm x=\pm y=\pm z$

Detailed Explanation

Theorem 1: Arithmetic Mean-Geometric Mean (or simply AM-GM) inequality for three variables

Let $a,b,c\in \mathbb R^+$, then $$\frac{a+b+c}{3}\ge\sqrt[3]{a\cdot b\cdot c}$$


Proof
Substitute $a=d^3, b=e^3, c=f^3$. Observe now that
\begin{align*}
&d^3+e^3+f^3-3\cdot def=\frac{1}{2}\cdot (d+e+f)\cdot \big((d-e)^2+(e-f)^2+(f-d)^2\big)\ge0\tag{1}\\
\iff & d^3+e^3+f^3\ge 3\cdot def\\
\iff & a+b+c\ge 3\cdot \sqrt[3]{abc}
\end{align*}
You can check the equation in $(1)$ here or prove it yourself - it's up to you.

Sidenote: Note that you'll only achive equality in $(1)$ if and only if $$d+e+f=0\qquad \text{or} \qquad (d-e)^2+(e-f)^2+(f-d)^2=0$$
i.e. if and only if $d=e=f\implies a=b=c$.

Now, back to your problem, observe that the function $f(x, y, z):=x^2\cdot y^2\cdot z^2$ can only take non-negative values as long as $f:\mathbb R\to \mathbb R^+$ - which I assume.
Thus - in virtue of the Theorem 1 $$\frac{289}{3}=\frac{x^2+y^2+z^2}{3}\ge \sqrt[3]{x^2y^2z^2} $$
