[m]\frac{ ∂ z}{ ∂ y}=(ln(x^2+xy+y^2))`_{y}=\frac{1}{x^2+xy+y^2}\cdot (x^2+xy+y^2)`_{y}=\frac{1}{x^2+xy+y^2}\cdot (x+2y)[/m]
[m] \frac{ dy}{ dx}=(\frac{1}{2}x^2+x)`_{x}=x+1[/m]
Тогда
[m]z`=\frac{ d z}{ d x}=\frac{ ∂ z}{ ∂ x}+\frac{ ∂ z}{ ∂y}\cdot \frac{ dy}{ dx}=\frac{1}{x^2+xy+y^2}\cdot (2x+y)+\frac{1}{x^2+xy+y^2}\cdot (x+2y)\cdot (x+1)[/m]