Calculating Gradient of a $2$ Variable Function Let $f(x,y)$ have continuous partial derivatives at every point. We know that
$$\nabla f(0,3)=5 {\bf{i}} - {\bf{j}}$$
Then we define $g(x,y)=f(x^2-y^2, 3x^2y)$.
I am not sure what I should do to calculate $\nabla g(-1,1)$. 
Any guidance would be appreciated.
 A: Hint: Recall the meaning of gradient (in 2D cartesian coordinates) $$\nabla f = \frac{\partial f}{\partial x}\mathbf i + \frac{\partial f}{\partial y}\mathbf j.$$ 
Also, use the chain rule for partial derivatives $$ \frac{\partial}{\partial x}F(G(x,y),H(x,y)) = \frac{\partial G}{\partial x} \frac{\partial F}{\partial G} + \frac{\partial H}{\partial x} \frac{\partial F}{\partial H} $$
A: You should use the chain rule! That is all! :)
Suppose that we have a scalar field $f(x,y)$ and a vector field ${\bf{v}}(x,y)=v_1(x,y) {\bf{i}} + v_2(x,y) {\bf{j}}$. Then we make a new scalar field by the composition of these as follows
$$g(x,y)=(f \circ {\bf{v}})(x,y) \equiv f(v_1(x,y),v_2(x,y))$$
Then let us use the chain rule to get
$$\begin{array}{}
\dfrac{\partial g}{\partial x} = (\dfrac{\partial f}{\partial x} \circ {\bf{v}}) \dfrac{\partial v_1}{\partial x}+(\dfrac{\partial f}{\partial y} \circ {\bf{v}}) \dfrac{\partial v_2}{\partial x} \\
\dfrac{\partial g}{\partial y} = (\dfrac{\partial f}{\partial x} \circ {\bf{v}}) \dfrac{\partial v_1}{\partial y}+(\dfrac{\partial f}{\partial y} \circ {\bf{v}}) \dfrac{\partial v_2}{\partial y}
\end{array}$$
or in a more compact notation we can write
$$\boxed{
\nabla g = \nabla ( f \circ {\bf{v}} ) =  \nabla {\bf{v}} \cdot (\nabla f \circ {\bf{v}})
}$$
or in matrix notation it is equivalent to
$$\begin{bmatrix}
\dfrac{\partial g}{\partial x} \\
\dfrac{\partial g}{\partial y}
\end{bmatrix}
=
\begin{bmatrix}
\dfrac{\partial v_1}{\partial x} & \dfrac{\partial v_2}{\partial x} \\
\dfrac{\partial v_1}{\partial y} & \dfrac{\partial v_2}{\partial y}
\end{bmatrix}
\begin{bmatrix}
\dfrac{\partial f}{\partial x} \circ {\bf{v}} \\
\dfrac{\partial f}{\partial y} \circ {\bf{v}}
\end{bmatrix}$$

Your Example
The vector field is
$${\bf{v}}(x,y)=(x^2-y^2) {\bf{i}} + (3x^2y) {\bf{j}}$$
and its value at $(-1,1)$ is
$${\bf{v}}(-1,1) = 0 {\bf{i}} + 3 {\bf{j}}$$
Then we compute $\nabla {\bf{v}}$ at the point $(-1,1)$ to get
$$\nabla {\bf{v}}(-1,1)
=
\begin{bmatrix}
2x & 6xy \\
-2y & 3x^2
\end{bmatrix}_{(-1,1)}
=
\begin{bmatrix}
-2 & -6 \\
-2 & 3
\end{bmatrix}
$$
and hence we have
$$\begin{align}
\nabla g(-1,1) &= \nabla {\bf{v}} (-1,1) \cdot (\nabla f \circ {\bf{v}}) (-1,1) \\
&= \nabla {\bf{v}} (-1,1) \cdot \nabla f (0,3)
\end{align}$$
What remains is just a matrix multiplication
$$
\nabla g(-1,1)
=
\begin{bmatrix}
-2 & -6 \\
-2 & 3
\end{bmatrix}
\begin{bmatrix}
5 \\
-1
\end{bmatrix}
=
\begin{bmatrix}
-4 \\
-13
\end{bmatrix}
$$
