The function is $f(x,y,z)=13x^4+13y^4+13z^4$, and the constraint is $13x^2+13y^2+13z^2=3$.
I've set $<f_x, f_y, f_z>=\lambda<g_x, g_y, g_z>$ and got:
$$<52x^3, 52y^3, 52z^3>=\lambda<26x, 26y, 26z>$$
Then I equated everything (assuming that $x$, $y$, or $z$ are not equal to 0) to get the three equations: $$x^2=\frac\lambda2 $$ $$y^2=\frac\lambda2$$ $$z^2=\frac\lambda2$$
After plugging these into the constraint I found $$3\left(\frac{13\lambda}{2}\right)=3$$ so $\lambda=\frac{2}{13}$
That must mean that $x$, $y$, and $z$ are all equal to $\sqrt\frac{1}{13}$
Plugging those into $f(x,y,z)$ gives $f=\left(\sqrt\frac{1}{13},\sqrt\frac{1}{13},\sqrt\frac{1}{13} \right)=\frac{3}{13}$, which should be the maximum value. But apparently the maximum value is $\frac{9}{13}$ and $x$, $y$, and $z$ all equal $\sqrt\frac{3}{13}$ . Can someone help me figure out where I went wrong?
As you found, a solution (there are eight) is $x=y=z=\frac{1}{\sqrt{13}}$.
All solutions are $$(x,y,z) \in \left\{ \left(\mp \frac{1}{\sqrt{13}},\mp \frac{1}{\sqrt{13}},\mp \frac{1}{\sqrt{13}} \right)\right\}$$
$$f(x,y,z) = 13(x^4+y^4+z^4) = 13 \cdot 3\cdot \frac{1}{13^2} = \frac{3}{13}.$$