$$6 \frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial x \partial y}-\frac{\partial^2 z}{\partial y^2}=1,$$

【答案】 $\frac{1}{25}$.

【解析】 由复合函数的链导法则, 有
\begin{aligned} &\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}=\frac{\partial z}{\partial u}+\frac{\partial z}{\partial v}, \\ &\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y}=-2 \frac{\partial z}{\partial u}+3 \frac{\partial z}{\partial v}, \end{aligned}

\begin{aligned} &\frac{\partial^2 z}{\partial x^2}=\frac{\partial^2 z}{\partial u^2}+2 \frac{\partial^2 z}{\partial u \partial v}+\frac{\partial^2 z}{\partial v^2}, \frac{\partial^2 z}{\partial x \partial y}=-2 \frac{\partial^2 z}{\partial u^2}+\frac{\partial^2 z}{\partial u \partial v}+3 \frac{\partial^2 z}{\partial v^2}, \\ &\frac{\partial^2 z}{\partial y^2}=4 \frac{\partial^2 z}{\partial u^2}-12 \frac{\partial^2 z}{\partial u \partial v}+9 \frac{\partial^2 z}{\partial v^2}, \end{aligned}