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$f(x,y)$ = $x^3 + y^3 - 3x^2 - 6y^2 - 9x$

Setting $fx$ and $fy$ equal to 0, I get $x = -1, 3$ and $y = 0, 4$. This is where I would normally plug in each x-value to find its corresponding y-value in the partial derivative, but neither partial derivative is in terms of both variables, so I'm not sure how to find the critical points (because, for example, I can't plug $x = -1$ to find what its y-value is).

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This just means that you have four equilibria points. From your derivatives you can construct the 2D gradient:

\begin{equation} \nabla f = f_x \mathbf{i} + f_y \mathbf{j}. \end{equation}

The function $f(x,y)$ has stationary points, $(x_e, y_e)$ whenever $\nabla f = \mathbf{0}$, which in your case occurs for the four points you have already calculated \begin{align} (x_e, y_e) &= (-1,0), \\ (x_e, y_e) &= (-1,4), \\ (x_e, y_e) &= (3,0), \\ (x_e, y_e) &= (3,4). \end{align}

And now you can just use your standard tests to determine the nature of these four equilibria.

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