Yes. The cubic polynomial $$ c(x,y) := 3x^3 + x^2y + y^3 + x^2 + 2xy + y^2 $$ has 3 isolated critical points at $(0,0)$, $(-\frac 6 {13}, - \frac{12}{13})$ and $(\frac 2 {19}, - \frac{8}{57})$.

My idea to find this example was to see the set of critical points at the intersection of two conics given by $c_x = 0$ and $c_y = 0$. If you want two conics to have 3 intersection points, one of them needs to be a tangent contact point. Hence, at this point, the gradient of $c_x$ and $c_y$ must be colinear. By translation, I assumed this double point to be $(0,0)$. This allowed by to set the coefficients of order $0$, $1$ and $2$ of $c$. I then chose the leading coefficients by trial and error.

For the example given above, $c_y = 0$ is an ellipse and $c_x = 0$ is an hyperbola.

Yes. The cubic polynomial $$ c(x,y) := 3x^3 + x^2y + y^3 + x^2 + 2xy + y^2 $$ has 3 isolated critical points at $(0,0)$, $(-\frac 6 {13}, - \frac{12}{13})$ and $(\frac 2 {19}, - \frac{8}{57})$.

The gradient is given by $$ \begin{align} c_x(x,y) & = 9x^2 + 2xy + 2x + 2y \\ c_y(x,y) & = x^2 + 3 y^2 + 2x + 2 y \end{align} $$

My idea to find this example was to see the set of critical points at the intersection of two conics given by $c_x = 0$ and $c_y = 0$. If you want two conics to have 3 intersection points, one of them needs to be a tangent contact point. Hence, at this point, the gradients of $c_x$ and $c_y$ must be colinear. By translation, I assumed this double point to be $(0,0)$. This allowed by to set the coefficients of order $0$, $1$ and $2$ of $c$. I then chose the leading coefficients by trial and error.

For the example given above, $c_y = 0$ is an ellipse and $c_x = 0$ is an hyperbola.

It is hard to tell on the graph of the two conics, but there are indeed two intersection points in the region where they overlap. You can visualize this better by expressing the curves $y_{red}(x)$ such that $c_y(x,y_{red}(x)) = 0$ and $y_{blue}(x)$ similarly and plotting $y_{red}-y_{blue}$.

Yes. The cubic polynomial $$ c(x,y) := 3x^3 + x^2y + y^3 + x^2 + 2xy + y^2 $$ has 3 isolated critical points at $(0,0)$, $(-\frac 6 {13}, - \frac{12}{13})$ and $(\frac 2 {19}, - \frac{8}{57})$.

Yes. The cubic polynomial $$ c(x,y) := 3x^3 + x^2y + y^3 + x^2 + 2xy + y^2 $$ has 3 isolated critical points at $(0,0)$, $(-\frac 6 {13}, - \frac{12}{13})$ and $(\frac 2 {19}, - \frac{8}{57})$.

My idea to find this example was to see the set of critical points at the intersection of two conics given by $c_x = 0$ and $c_y = 0$. If you want two conics to have 3 intersection points, one of them needs to be a tangent contact point. Hence, at this point, the gradient of $c_x$ and $c_y$ must be colinear. By translation, I assumed this double point to be $(0,0)$. This allowed by to set the coefficients of order $0$, $1$ and $2$ of $c$. I then chose the leading coefficients by trial and error.

For the example given above, $c_y = 0$ is an ellipse and $c_x = 0$ is an hyperbola.