已知函数f(u)f(u) 在区间(0,+∞)\begin{pmatrix}0,+\infty\end{pmatrix} 内具有二阶导数,记g(x,y)=f=),若g(x,y)g(x,y) 满足x2x^{2} ∂2g∂x2+xy\frac {\partial ^{2}g}{\partial x^{2}}+ xy ∂2g∂x∂y+y2\frac {\partial ^{2}g}{\partial x\partial y}+ y^{2} ∂2g∂y2=1\frac {\partial ^{2}g}{\partial y^{2}}= 1 ,且g(x,x)=1g(x,x)=1 · ∂g∂x∣(x,x)=2x\left.\frac{\partial g}{\partial x}\right|_{(x,x)}=\frac2x ,求f(u)f(u)
f(u)=12(lnu)2+2lnu+1f(u)=\frac{1}{2}\left(\ln u\right)^2+2\ln u+1
3