Check where the first-order partial derivatives vanish to find any critical points within the given region:
[tex]f(x,y)=2x^3+y^4\implies\begin{cases}f_x=6x^2=0\\f_y=4y^3=0\end{cases}\implies(x,y)=(0,0)[/tex]
The Hessian for this function is
[tex]\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}12x&0\\0&12y^2\end{bmatrix}[/tex]
with [tex]\det\mathbf H(0,0)=0[/tex], so unfortunately the second partial derivative test fails. However, if we take [tex]x=0[/tex] we see that [tex]f(x,y)>0[/tex] for different values of [tex]y[/tex]; if we take [tex]y=0[/tex] we see [tex]f(x,y)[/tex] takes on both positive and negative values. This indicates (0, 0) is neither the site of an extremum nor a saddle point.
Now check for points along the boundary. We can parameterize the boundary by
[tex](x,y)=(8\cos t,8\sin t)[/tex]
with [tex]0\le t\le2\pi[/tex]. This turns [tex]f(x,y)[/tex] into a univariate function [tex]F(t)[/tex]:
[tex]F(t)=f(8\cos t,8\sin t)=2^{10}\cos^3t+2^{12}\sin^4t[/tex]
[tex]\implies F'(t)=3\cdot2^{10}\cos^2t(-\sin t)+2^{14}\sin^3t\cos t=2^{10}\sin t\cos t(16\sin^2t-3\cos t)[/tex]
[tex]F'(t)=0\implies\begin{matrix}\sin t=0\implies t=0,\,t=\pi\\\\\cos t=0\implies t=\dfrac\pi2,\,t=\dfrac{3\pi}2\\\\16\sin^2t-3\cos t=0\implies t=2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3},\,t=2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\end{matrix}[/tex]
At these critical points, we get
[tex]F(0)=1024[/tex]
[tex]F\left(2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893[/tex]
[tex]F\left(\dfrac\pi2\right)=4096[/tex]
[tex]F(\pi)=-1024[/tex]
[tex]F\left(\dfrac{3\pi}2\right)=4096[/tex]
[tex]F\left(2\pi-2\tan^{-1}\sqrt{\dfrac{\sqrt{1033}-32}3}\right)\approx893[/tex]
We only care about 3 of these results.
[tex]t=\dfrac\pi2\implies(x,y)=(0,8)[/tex]
[tex]t=\pi\implies(x,y)=(-8,0)[/tex]
[tex]t=\dfrac{3\pi}2\implies(x,y)=(0,-8)[/tex]
So to recap, we found that [tex]f(x,y)[/tex] attains
