Derivative of a composite function $h = f\circ g$ where $f\colon \mathbb{R}^2\to \mathbb{R}$

Question

$f\colon \mathbb{R}^2\to \mathbb{R}$ such that $f_x(2,-1)=1$ and $f_y(2,-1)=1$ and $g(x,y)=\langle x^2y,x-y\rangle$ and $h = f\circ g$ then find $h_y(1,2)$. The options given are $-2,2,0,5,-5,-3,10,3$

My Attempt

$$ h_y(1,2)=(f\circ g)_y(x,y)\Big|_{x=1,y=2}=f_y(g(1,2))\cdot g_y(1,2)=f_y(2,-1)\cdot g_y(1,2)\\=1 \cdot g_y(1,2) $$

Is it the right way towards finding the solution ? How do I find $g_y(1,2)$ ?

Answer 1

If $\langle \cdot,\cdot \rangle$ defines a vector, then you did it wrong. Using the chain rule (the case when $k=1$) we have

$$ h_{y}(x,y) =(\nabla f)(g(x,y)) \cdot g_{y}(x,y)=\langle f_{x}(g(x,y)),f_{y}(g(x,y)) \rangle \cdot \langle x^{2},-1 \rangle.$$

Thus

\begin{align*} h_{y}(1,2) &=\langle f_{x}(g(1,2)),f_{y}(g(1,2)) \rangle \cdot \langle 1,-1 \rangle \\ &= \langle f_{x}(2,-1),f_{y}(2,-1) \rangle \cdot \langle 1,-1 \rangle \\ &= \langle 1,1 \rangle \cdot \langle 1,-1 \rangle \\ &=0 \end{align*}

Answer 2

Since $f$ is a multivariable function, and $g$ is a vector function, we can write it as

$$\frac{\partial h}{\partial y} = \vec{\nabla f} . \vec{\frac{\partial g}{\partial y}}$$