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This may sound stupid but I am really confused.

Let's assume $z=g(x)$ and $y=f(x)$. The derivative of $y$ with respect to $x$ is $f'(x)$. Suppose $f'(x^*)=0$. So, $z$ at $x^{*}$ is $g(x^{*})$.

Now let's compute ${\partial z\over\partial y}$,

$${\partial z\over\partial y}=g'(x){\partial x\over\partial y}$$

$${\partial z\over\partial y}=g'(x)\frac{1}{\partial y\over\partial x}$$

So, ${\partial z\over\partial y}$ is undefined at $x^*$. That means $z$ is undefined at $x^*$ but previously we saw that $z$ at $x^{*}$ is $g(x^{*})$. What is the justification?

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    I don't understand: why do you think $\;z\;$ is a function of $\;y\;$ ? If it isn't then simply $\;\cfrac{dz}{dy}=0\;$ ...And even if this isn't a problem, why the derivative $\;z'_y\;$ not existing at $\;x^*\;$ would imply $\;z\;$ isn't defined there? – 2017-01-02
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    I thought we can write $z=g(f^{-1}(y))$. But we know that $z$ is defined at $x^*$ – 2017-01-02
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    You can write $z=g(f^{-1}(y))$ but only when $f$ is invertible. One sufficient condition for the same is that $f'(x) \neq 0$. – 2017-01-02
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    Also note that the derivative $dz/dy$ may exist even if $dy/dx=0$. Thus if $z=x^{6},y=x^{3}$ then $z=y^{2}$ and $dz/dy=0$at $y=0$ even though $dy/dx=0$ at $x=y=0$. – 2017-01-02
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    Also why do you use the symbol $\partial$ which is normally needed for partial derivatives? – 2017-01-02
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    @Paramanand Singh, yes it should be $dy$ instead of $\partial{y}$. Regarding your first comment, consider $z=x^3,y=x^2$. Then we can write $z=y^{3/2}$ but $f'(x)=0$ at $x=0$. – 2017-01-02
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    The condition $f'(x) \neq 0$ is sufficient but not necessary. The right example is given in my previous comment. Your example of $z=y^{3/2}$ has an issue because if $x<0$ then $y>0,z<0$ but $z=y^{3/2}$ implies that both $y, z$ are non-negative. – 2017-01-02

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You have discovered the hard part of proving the chain rule for the derivative of a composite function.

If $h(x) = f(g(x))$, the chain rule states that $h'(x) = g'(x)f'(g(x))$.

The proof of this is easy at any $x$ such that $g'(x) \ne 0$.

However, if $g'(x) = 0$, the proof is more difficult, and requires careful reasoning.

So, congratulations.

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So, $\frac{\partial z}{\partial y}$ is undefined at $x^*$. That means $z$ is undefined at $x^*$ but previously we saw that $z$ at $x^{*}$ is $g(x^{*})$. What is the justification?

As noted in the comments by Paramanand Singh, there are technical issues surrounding the possibility of inverting $f$ to obtain $z = g(f^{-1}(y))$ from $z = g(x)$ and $y = f(x)$.

Even if inverting $f$ is possible, however, just because a function's derivative is undefined at a point does not mean the function value is undefined.

For example, take $z = g(x) = x$ and $y = f(x) = x^{3}$ (and $x^{*} = 0$), so that $z = g(f^{-1}(y)) = y^{1/3}$.