Calculating the first order partial derivatives of $z(x,y.)$.

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My function is $z^3 - 3xyz = 1$ and I calculated $z_{x}^{'}$ and I got $z_{x}^{'} = \frac{yz + yy^{'}z}{z^{2} - xy }$. but the answer at the back of the book is $z_{x}^{'} = \frac{yz}{z^{2} - xy },$ could anyone clarify for me if I am wrong and why?

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We get by the chain rule $$3z^2z_x-3yz-3xyz_x=0$$ so $$z_x(z^2-xy)=yz$$ Hint: By the quotient rule we obtain $$z_{xx}=\frac{yz_x(z^2-xy)-yz(2zz_x)}{(z^2-xy)^2}$$ and for $$z_x$$ you must plug in $$z_x=\frac{yz}{z^2-xy}$$