Use definition to prove that the function $f(x,y)=xye^{xy}$ is differentiable at all points in $\mathbb{R}^2$
I tried to solve this question like so:-
Let $(x_0,y_0)$ be any point in $\mathbb R^2$
$$\begin{aligned} f_x&=y(e^{xy}+xe^{xy}y) \qquad\text{and} \qquad &f_{y}&=x(e^{xy}+ye^{xy}x)\\ &=ye^{xy}+xy^2e^{xy} & &=xe^{xy}+x^2ye^{xy}\end{aligned}$$
Therefore,
$$\begin{aligned} f_x(x_0,y_0)=y_0e^{x_0y_0}+x_0y_0^2e^{x_0y_0} \qquad\text{and} \qquad f_y(x_0,y_0)=x_0e^{x_0y_0}+x_0^2y_0e^{x_0y_0}\end{aligned}$$
Hence,
$$\begin{aligned}\Delta f(x_0,y_0)&=f(x_0+\Delta x,y_0+\Delta y) - f(x_0,y_0)\\ &=(x_0y_0+x_0\Delta y + y_0\Delta x+\Delta x\Delta y)e^{x_0y_0+x_0\Delta y + y_0\Delta x+\Delta x\Delta y}-x_0y_0e^{x_0y_0}\qquad\small{\text{(By Substitution)}}\end{aligned}$$
And hence, I am not able to form them as
$$\Delta f(x_0,y_0)=f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y+\epsilon_1\Delta x+\epsilon_2\Delta y$$
Sorry if it's really simple but I cannot figure a way how I can do it.
The definition mentioned in the question is as below:-
If $f$ is a function of two variables $x$ and $y$ and the increment of $f$ at $(a,b)$ $\left(\text{i.e.} \Delta f(a,b)=f_x(a,b)\Delta x + f_y(a,b)\Delta y + \epsilon_1\Delta x + \epsilon_2\Delta y\right)$ where $\epsilon_1$ and $\epsilon_2$are functions of $\Delta x$ and $\Delta y$ such that $\epsilon_1 \to 0$ and $\epsilon_2\to0$ as $(\Delta x,\Delta y)\to(0,0)$, then function $f$ is said to be $\underline{\text{differentiable}}$ at $(a,b).$
