已知函数 f(u,v) 具有二阶连续偏导数，且函数 g(x,y)=f(2x+y,3x-y) 满足 (partial^2 g)/(partial x^2)+(partial^2 g)/(partial xpartial y)-6(partial^2 g)/(partial y^2)=1.(1) 求 (partial^2 f)/(partial upartial v)；(2) 若 (partial f(u,0))/(partial u)=ue^-u，f(0,v)=(1)/(50)v^2-1，求 f(u,v) 的表达式.

(1) 由 $ g(x, y) = f(2x + y, 3x - y) $，利用链式法则计算二阶偏导数，得 \[ \frac{\partial^2 g}{\partial x^2} = 4f_{uu} + 12f_{uv} + 9f_{vv}, \quad \frac{\partial^2 g}{\partial x \partial y} = 2f_{uu} + f_{uv} - 3f_{vv}, \quad \frac{\partial^2 g}{\partial y^2} = f_{uu} - 2f_{uv} + f_{vv} \] 代入条件 $\frac{\partial^2 g}{\partial x^2} + \frac{\partial^2 g}{\partial x \partial y} - 6 \frac{\partial^2 g}{\partial y^2} = 1$，化简得 $25f_{uv} = 1$，故 \[ \boxed{\frac{1}{25}} \] (2) 由 $ f_{uv} = \frac{1}{25} $，积分得 $ f_u(u, v) = \frac{v}{25} + g(u) $，其中 $ g(u) = ue^{-u} $。再积分得 \[ f(u, v) = \frac{vu}{25} - e^{-u}(u + 1) + h(v) \] 由 $ f(0, v) = \frac{1}{50}v^2 - 1 $，得 $ h(v) = \frac{1}{50}v^2 $。因此， \[ \boxed{\frac{v^2}{50} + \frac{uv}{25} - (u + 1)e^{-u}} \]