解：

令\( u = \frac{x}{y} \)，则\( \frac{\partial g}{\partial x} = f'(u)\frac{1}{y} \)，\( \frac{\partial g}{\partial y} = f'(u)\left(-\frac{x}{y^2}\right) \)

又\( g(x, x) = f\left(\frac{x}{x}\right) = f(1) = 1 \)，\(\left.\frac{\partial g}{\partial x}\right\rvert_{(x, x)} = f'(1)\frac{1}{x} = \frac{2}{x} \)，故\( f'(1) = 2 \)

\( \frac{\partial^2 g}{\partial x^2} = \left(f''(u)\frac{1}{y}\right)\frac{1}{y} = f''(u)\frac{1}{y^2} \)  ......(1)

\( \frac{\partial^2 g}{\partial x\partial y} = f''(u)\left(-\frac{x}{y^2}\right)\frac{1}{y} + f'(u)\left(-\frac{1}{y^2}\right) = -\frac{x}{y^3}f''(u) - \frac{1}{y^2}f'(u) \)  ..(2)

\( \frac{\partial^2 g}{\partial y^2} = f''(u)\left(-\frac{x}{y^2}\right)\frac{x}{y} + f'(u)\left(\frac{2x}{y^2}\right) = \frac{x^2}{y^4}f''(u) + \frac{2x}{y^3}f'(u) \)  ...(3)

将（1）（2）（3）代入\( x^2\frac{\partial^2 g}{\partial x^2} + xy\frac{\partial^2 g}{\partial x\partial y} + y^2\frac{\partial^2 g}{\partial y^2} = 1 \)化简得：

\( u^2f''(u) + uf'(u) = 1 \)，即\( f''(u) + \frac{1}{u^2}f'(u) = \frac{1}{u^2} \)。

令\( p' = f'(u) \)则\( p'' + \frac{1}{u}p' = \frac{1}{u^2} \)

\( p = e^{-\int \frac{1}{u}du}\left[\int \frac{1}{u^2}e^{\int \frac{1}{u}du}du + C\right] = \frac{1}{u}\left[\int \frac{1}{u}du + C\right] = \frac{\ln u}{u} + \frac{C}{u} \)

又\(\left.p'\right\rvert_{u = 1} = C = 2\)，故\( p = \frac{\ln u}{u} + \frac{2}{u} \)

因此，\( \frac{\partial f}{\partial u} = \frac{\ln u}{u} + \frac{2}{u} \)，积分得\( f(u) = \int \left(\frac{\ln u}{u} + \frac{2}{u}\right)du = \frac{1}{2}\ln^2 u + 2\ln u + C \)，

又\( f(1) = C = 1 \)，故\( f(u) = \frac{1}{2}\ln^2 u + 2\ln u + 1 \) 。