If $z=f(x,y)$ and $y=x^2$, then by the chain rule
$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial z}{\partial x}+2x\frac{\partial z}{\partial y}$
Therefore
$2x\frac{\partial z}{\partial y}=0$
and
$\frac{\partial z}{\partial y}=0$
What is wrong with this argument?
I have a feeling that
1.) $x$ and $y$ do not have partial derivatives because they are single-variable, and
2.) $\frac{\partial z}{\partial y}$ cannot be zero, because $y=x^2$ and therefore the derivative of any $y$ term exists.
How is my reasoning? I am pretty confused by this question.
Nothing wrong. Just change it into
$$\frac{d z}{d x}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial z}{\partial x}+2x\frac{\partial z}{\partial y}$$
Note that that the first term is $\frac{d z}{d x}$, which is different from $\frac{\partial z}{\partial x}$. So they cannot cancel out. The partial derivatives do exist, but be careful not to mix it up with $\frac{d z}{d x}$, which is NOT a partial derivative.
Actually, a better way to say this is that
$$\left[\frac{\partial z}{\partial x}\right]_{y=x^2}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial z}{\partial x}+2x\frac{\partial z}{\partial y}.$$
Where I have clearly written down the restriction $y=x^2$.