(2) $\frac{\partial f}{\partial x}(x, y), \frac{\partial f}{\partial y}(x, y)$ 在点 $(0,0)$ 处不连续;
(3) $f(x, y)$ 在 $(0,0)$ 处不可微.

(1)证明:
\begin{aligned} & \frac{\partial f}{\partial x}(0,0)=\lim _{\substack{x \rightarrow 0 \\ y=0}} \frac{f(x, 0)-f(0,0)}{x-0}=\lim _{\substack{x \rightarrow 0 \\ y=0}} \frac{f(x, 0)-f(0,0)}{x}=\lim _{\substack{x \rightarrow 0 \\ y=0}} \frac{0-0}{x}=0 ; \\ & \frac{\partial f}{\partial y}(0,0)=\lim _{\substack{x=0 \\ y \rightarrow 0}} \frac{f(x, 0)-f(0,0)}{x-0}=\lim _{\substack{x=0 \\ y \rightarrow 0}} \frac{f(0, y)-f(0,0)}{y}=\lim _{\substack{x=0 \\ y \rightarrow 0}} \frac{0-0}{y}=0 ; \end{aligned}

(2)证明:

\begin{aligned} & \frac{\partial f}{\partial x}=\frac{2 x y \cdot\left(x^4+y^2\right)-x^2 y \cdot 4 x^3}{\left(x^4+y^2\right)^2}=\frac{2 x^5 y+2 x y^3-4 x^5 y}{\left(x^4+y^2\right)^2}=\frac{2 x y^3-2 x^5 y}{\left(x^4+y^2\right)^2} ; \\ & \frac{\partial f}{\partial y}=\frac{x^2 \cdot\left(x^4+y^2\right)-x^2 y \cdot 2 y}{\left(x^4+y^2\right)^2}=\frac{x^6+x^2 y^2-2 x^2 y^2}{\left(x^4+y^2\right)^2}=\frac{x^6-x^2 y^2}{\left(x^4+y^2\right)^2} ; \end{aligned}

$$f_x(x, y)=\left\{\begin{array}{ll} \frac{2 x y^3-2 x^5 y}{\left(x^4+y^2\right)^2}, & x^2+y^2 \neq 0 \\ 0, & x^2+y^2=0 \end{array} \text {, 选取 } y=2 x^2\right. \text { 时: }$$

\begin{aligned} & \lim _{\substack{x \rightarrow 0 \\ y=2 x^2}} f_x(x, y)=\lim _{\substack{x \rightarrow 0 \\ y=2 x^2}} \frac{2 x y^3-2 x^5 y}{\left(x^4+y^2\right)^2}=\lim _{x \rightarrow 0} \frac{2 x \cdot\left(2 x^2\right)^3-2 x^5 \cdot 2 x^2}{\left[x^4+\left(2 x^2\right)^2\right]^2}=\lim _{x \rightarrow 0} \frac{2 x \cdot 8 x^7-4 x^7}{\left(x^4+4 x^4\right)^2} \\ & =\lim _{x \rightarrow 0} \frac{12 x^7}{\left(5 x^4\right)^2}=\lim _{x \rightarrow 0} \frac{12 x^7}{25 x^8}=\frac{12}{25} \lim _{x \rightarrow 0} \frac{1}{x}=\infty, \text { 极限不存在; } \\ & f_y(x, y)=\left\{\begin{array}{l} \frac{x^6-x^2 y^2}{\left(x^4+y^2\right)^2}, x^2+y^2 \neq 0 \\ 0, \end{array}, \text { 选取取 } y=2 x^2\right. \text { 时: } \\ & \lim _{\substack{x \rightarrow 0 \\ y=2 x^2}} f_y(x, y)=\lim _{x \rightarrow 0} \frac{x^6-x^2 y^2}{\left(x^4+y^2\right)^2}=\lim _{x \rightarrow 0} \frac{x^6-x^2 \cdot\left(2 x^2\right)^2}{\left[x^4+\left(2 x^2\right)^2\right]^2}=\lim _{x \rightarrow 0} \frac{-3 x^6}{25 x^8} \\ & =-\frac{3}{25} \lim _{x \rightarrow 0} \frac{1}{x^2}=\infty, \text { 极限不存在; } \end{aligned}

(3)证明 : 二重极限存在 $\Longleftrightarrow$ 关于收敛点距离一致收玫

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