I am trying to prove chain rule for Partial derivatives:
Let $z=f(x,y)$ and $x=x(t),y(t)$. We need to prove that: $$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial t}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial t}$$
Assuming $z$ is differentiable function we have the increment formula as: $$\Delta z=f_x\Delta x+f_y \Delta y+\epsilon_1 \Delta x+\epsilon_2 \Delta y$$
Where $\epsilon_1,\epsilon_2 \to 0$ as $(\Delta x,\Delta y) \to (0,0)$
Now we have: $$\frac{\Delta z}{\Delta t}=f_{x} \frac{\Delta x}{\Delta t}+f_{y} \frac{\Delta y}{\Delta t}+\varepsilon_{1} \frac{\Delta_{x}}{\Delta t}+\varepsilon_{2} \frac{\Delta y}{\Delta t}$$
Taking $\Delta t \to 0$ we get: $$\lim _{\Delta t \rightarrow 0} \frac{\Delta z}{\Delta t}=f_{x} \lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}+f_{y} \lim _{\Delta t \rightarrow 0} \frac{\Delta y}{\Delta t}+\varepsilon_{1} \lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}+\varepsilon_{2} \lim _{\Delta t \rightarrow 0} \frac{\Delta y}{\Delta t}$$
Any way to prove from here?