Question

if z^3-xz-y=0 prove d^2z/dxdy=(-3z^2+x)/(3z^2-x)^3

Answer 1

\[z^3-xz-y=0\]\[3z^2z'_x-z-xz'_x=0\quad\Rightarrow\quad z'_x=\frac{z}{3z^2-x}\quad(1)\]\[3z^2z'_y-xz'_y-1=0\quad\Rightarrow\quad z'_y=\frac{1}{3z^2-x}\quad(2)\]\[(2)\Rightarrow\quad 6zz'_xz'_y+3z^2z''_{xy}-z'_y-xz''_{xy}=0\]\[z''_{xy}=\frac{(1-6zz'_x)z'_y}{3z^2-x}=\]\[=\frac{1}{3z^2-x}\cdot\left(1-6z\frac{z}{3z^2-x}\right)\cdot\frac{1}{3z^2-x}=\]\[=\frac{-3z^2-x}{(3z^2-x)^3}\]