편도함수

As $z(x,y)=f(x^2+y^2)$

$$\begin{aligned} \frac{\partial z}{\partial x}(x,y)&=2xf'(x^2+y^2) \\ \frac{\partial z}{\partial y}(x,y)&=2yf'(x^2+y^2) \end{aligned}$$

by the (ordinary single variable) chain rule. So

$$y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y} =y(2x)f'(x^2+y^2)-x(2y)f'(x^2+y^2)=0$$

and the differential equation is always satisfied, assuming that $f$ is differentiable, so that the chain rule applies.

By definition

$$\begin{aligned} \frac{\partial f}{\partial x}(0,0) &=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)-f(0,0)}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}\frac{\frac{(\Delta x+2\times 0)^2}{\Delta x+0}-0}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}\frac{\Delta x}{\Delta x} \\ &=1 \end{aligned}$$

and

$$\begin{aligned} \frac{\partial f}{\partial y}(0,0) &=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)-f(0,0)}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}\frac{\frac{(0+2\Delta y)^2}{0+\Delta y}-0}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}\frac{4\Delta y}{\Delta y} \\ &=4 \end{aligned}$$

(b) $f(x,y)$ is not continuous at $(0,0)$, even though both partial derivatives exist there. To see this, make a change of coordinates from $(x,y)$ to $(X,y)$ with $X=x+y$ (the denominator). Of course, $(x,y)\rightarrow (0,0)$ if and only if $(X,y)\rightarrow (0,0)$. Now watch what happens when $(X,y)\rightarrow(0,0)$ with $X$ a lot smaller than $y$. For example, $X=ay^2$. Then

$$\begin{aligned} \frac{(x+2y)^2}{x+y}=\frac{(X+y)^2}{X}=\frac{(ay^2+y)^2}{ay^2} =\frac{(1+ay)^2}{a}\rightarrow\frac{1}{a} \end{aligned}$$

This depends on $a$. So approaching $(0,0)$ along different paths gives different limits. (You can see the same effect without changing coordinates by sending $(x,y)\rightarrow (0,0)$ with $x=-y+ay^2$.) Even more dramatically, watch what happens when $(X,y)\rightarrow(0,0)$ with $X=y^3$. Then

$$\frac{(x+2y)^2}{x+y}=\frac{(X+y)^2}{X}=\frac{(y^3+y)^2}{y^3} =\frac{{(1+y^2)}^2}{y}\rightarrow\pm\infty$$