Can anybody please help me out, i can't figure out how to work this one
$$3y=z^3+3xz$$ $$0=3z^2\frac{\delta z}{\delta x}+3z+3x\frac{\delta z}{\delta x}$$ $$0=z^2\frac{\delta z}{\delta x}+z+x\frac{\delta z}{\delta x}$$ $$z=-(z^2+x)\frac{\delta z}{\delta x}$$
from $$0=z^2\frac{\delta z}{\delta x}+z+x\frac{\delta z}{\delta x}$$ $$0=2z(\frac{\delta z}{\delta x})^2+z^2\frac{\delta^2 z}{\delta x^2}+\frac{\delta z}{\delta x}+\frac{\delta z}{\delta x}+x\frac{\delta^2 z}{\delta x^2}$$ $$\frac{\delta^2 z}{\delta x^2}(z^2+x)=-2(z(\frac{\delta z}{\delta x})^2+\frac{\delta z}{\delta x})$$ taking in account that $$\frac{\delta z}{\delta x}=\frac{-z}{z^2+x}$$ follow that $$\frac{\delta^2 z}{\delta x^2}=\frac{2xz}{(z^2+x)^3}$$
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$$3y=z^3+3xz$$ $$0=3z^2\frac{\delta z}{\delta x}+3z+3x\frac{\delta z}{\delta x}$$ $$0=z^2\frac{\delta z}{\delta x}+z+x\frac{\delta z}{\delta x}$$ $$z=-(z^2+x)\frac{\delta z}{\delta x}$$
from $$0=z^2\frac{\delta z}{\delta x}+z+x\frac{\delta z}{\delta x}$$ $$0=2z(\frac{\delta z}{\delta x})^2+z^2\frac{\delta^2 z}{\delta x^2}+\frac{\delta z}{\delta x}+\frac{\delta z}{\delta x}+x\frac{\delta^2 z}{\delta x^2}$$ $$\frac{\delta^2 z}{\delta x^2}(z^2+x)=-2(z(\frac{\delta z}{\delta x})^2+\frac{\delta z}{\delta x})$$ taking in account that $$\frac{\delta z}{\delta x}=\frac{-z}{z^2+x}$$ follow that $$\frac{\delta^2 z}{\delta x^2}=\frac{2xz}{(z^2+x)^3}$$