Compute the gradient of the following function and evaluate it at the given point $\mathrm{P}$.
\[
g(x, y)=x^{2}-7 x^{2} y-6 x y^{2} ; P(-1,3)
\]
The gradient is $\nabla f(x, y)=$
The gradient at $(-1,3)$ is

Step 1 ：The gradient of a function $f(x, y)$ is given by the vector of its partial derivatives, i.e., $\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$.

Step 2 ：First, we compute the partial derivative of $g(x, y)$ with respect to $x$. Using the power rule and the product rule, we get $\frac{\partial g}{\partial x} = 2x - 14xy - 6y^2$.

Step 3 ：Next, we compute the partial derivative of $g(x, y)$ with respect to $y$. Again using the power rule and the product rule, we get $\frac{\partial g}{\partial y} = -7x^2 - 12xy$.

Step 4 ：Therefore, the gradient of $g(x, y)$ is $\nabla g(x, y) = \left(2x - 14xy - 6y^2, -7x^2 - 12xy\right)$.

Step 5 ：To evaluate the gradient at the point $P(-1,3)$, we substitute $x = -1$ and $y = 3$ into the gradient. We get $\nabla g(-1, 3) = \left(2(-1) - 14(-1)(3) - 6(3)^2, -7(-1)^2 - 12(-1)(3)\right)$.

Step 6 ：Simplifying the above expression, we get $\nabla g(-1, 3) = \left(-2 + 42 - 54, -7 + 36\right) = \left(-14, 29\right)$.

Step 7 ：So, the gradient of the function $g(x, y)$ at the point $P(-1,3)$ is $\boxed{(-14, 29)}$.