편도함수

If $f_y(0,0)<0$, then $f(0,y)$ decreases as $y$ increases from $0$. Thus moving in the positive $y$ direction takes you downhill. This means you aren't at the lowest point in a valley, since you can still move downhill. On the other hand, as $f_y(0,0)<0$, $f(0,y)$ also decreases as $y$ increases towards $0$ from slightly negative values. Thus if you move in the negative $y$-direction from $y=0$, your height $z$ will increase. So you are not at a locally highest point—you're not at a summit.

(a) By definition

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

(b) By definition

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

(c) By definition

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} f(t,t)\Big|_{t=0} &=\lim_{t\rightarrow 0}\frac{f(t,t)-f(0,0)}{t} \\ &=\lim_{t\rightarrow 0}\frac{\frac{(t^2)(t)}{t^2+t^2}-0}{t} \\ &=\lim_{t\rightarrow 0}\frac{t/2}{t} \\ &=\frac{1}{2} \end{aligned}$$

No such $f(x,y)$ exists, because if it were to exist, then we would have that $f_{xy}(x,y)=f_{yx}(x,y)$. But

$$\begin{aligned} f_{xy}(x,y)&=\frac{\partial }{\partial y}f_x(x,y)=\frac{\partial }{\partial y}e^y=e^y \\ f_{yx}(x,y)&=\frac{\partial }{\partial x}f_y(x,y)=\frac{\partial }{\partial x}e^x=e^x \end{aligned}$$

are not equal.