You are given $\nabla f(0, 1) = (3, 4)$, let $h(x, y) = f(y - x - 1, y - x)$, what is $\nabla h(4, 3)?$
To my understanding cant we just do
$$\nabla h(4, 3) = \nabla f(0, 1) = (3, 4)$$
Or do we have to go through the process of finding $h_x(0, 1)$ and $h_y(0, 1)$?
On
As you said, by definition $$ \nabla h(x_0,y_)) = \begin{pmatrix} h_x(x_0,y_0) \\ h_y(x_0,y_0) \end{pmatrix}. $$ In your case, for instance, $$ h_x(x_0,y_0)= f_x(x_0,y_0) \frac{\partial (y-x-1)}{\partial x} + f_y(x_0,y_0) \frac{\partial (y-x)}{\partial y}. $$ Of course all derivatives must be evaluated at $(x_0,y_0)$. Can you conclude?
You just need to use the chain rule appropriately. I have seen many fellas who become confused when things come to chain rule. My experience tells me that the confusion is mostly due to inappropriate notations that are used.
Note that $f:\mathbb{R}^2\to\mathbb{R},\,\nabla f:\mathbb{R}^2\to\mathbb{R}^2$. Also, we have a function which is not defined explicitly in the question, i.e.
\begin{align*} g\,:&\mathbb{R}^2\to\mathbb{R}^2 \\ &(x,y)\to(y-x-1,y-x) \end{align*}
next, we have the following
$$h=f \circ g$$
Consequently, by the chain rule we have
$$\nabla h=\nabla[f\circ g]=\Big[[\nabla f]\circ g\Big]\cdot\nabla g$$
when we write the above functional equation at $(x,y)$, we get
\begin{align*} \nabla h(x,y)&=\Big[[\nabla f]\circ g\Big](x,y)\cdot\nabla g(x,y) \\ &=\nabla f\big(g(x,y)\big)\cdot\nabla g(x,y) \\ &=\nabla f\big(y-x-1,y-x\big)\cdot\nabla g(x,y) \end{align*}
I think you can proceed now.