The question I'm trying to solve is
Find the linear approximation to the function defined by $f(x, y, z) = (zy^3, xe^x +y +z)$ at the point $(0, 1, 1)$.
If this was $\mathbb R^3 \to \mathbb R$ then I would get the gradient and from that get the equation for the tangent plane. Does the same process apply here? If so, how do I get the gradient for $\mathbb R^3 \to\mathbb R^2$?
The linear approximation is $$L(x,y,z)=f(0,1,1)+f'(0,1,1)\,(x,y,z)^T,$$where $f'$ is the (Fréchet) derivative $$ f'(a,b,c)=\begin{bmatrix}\frac{\partial f_1}{\partial x}(a,b,c)&\frac{\partial f_1}{\partial y}(a,b,c)&\frac{\partial f_1}{\partial z}(a,b,c) \\ \frac{\partial f_2}{\partial x}(a,b,c)&\frac{\partial f_2}{\partial y}(a,b,c)&\frac{\partial f_2}{\partial z}(a,b,c) \end{bmatrix}. $$ This is the unique linear transformation $T$ such that $$ \lim_{h\to0}\frac{\|f((a,b,c)+h)-f(a,b,c)-Th\|}{\|h\|}=0. $$