Critical values of $f(x, y) = x^3 − 147xy + 343y^3$ The only critical values I can find are $(0,0$) and $(7,1$). Apparently there is another one, and I can't find it which is very concerning. Just so we are clear, critical values are where $f_x = 0$ and $f_y = 0$ or either one DNE.
$$f(x, y) = x^3 − 147xy + 343y^3$$

Partial for $x$:

$$\frac{\partial f}{\partial x}=3x^2-147y$$

Partial for $y$:

$$\frac{\partial f}{\partial y}=-147x+1029y^2$$

Critical points are when $\frac{\partial f}{\partial x}$ AND $\frac{\partial f}{\partial y}$ are equal to $0$. Hence:

$$3x^2-147y=0$$

$$1029y^2-147x=0$$

We can solve for one variable, let's just solve for $x$:

$$1029y^2=147x$$
$$x=7y^2$$

Now use it:

$$3(7y^2)^2-147y=0$$
$$147y^4-147y=0$$
$$147y(y^3-1)=0$$

Difference of cubes:

$$147y(y^3-1)=147y(y-1)(y^2+y+1)=0$$

Now we can find the three zeroes:

$$147y=0$$
$$y-1=0$$
$$y^2+y+1=0$$

We get that $y=0, y=1, y=\frac{-1\pm 3i}{2}$. From those, you can solve for $x$ by plugging back in.
I am down to my last try. I don't know how to solve for the last point. I got two points which where $x=0$ and $x=7$.