Is the partial derivative of a constant always zero? I'm trying to get my head round using the multivariable chain rule to find exact derivatives. For example I want to find the exact derivative(using partial derivatives) of, 
$$r^2=x^2+y^2$$
Where r is initially a constant. I now assign $f(x,y)=r^2$, then it follows
$$\frac{\partial f}{\partial x}=2x, \frac{\partial f}{\partial y}=2y$$
Then from the multivariable chain rule,
$$\frac{df}{dx}=\frac{\partial f}{\partial x}*1 + \frac{\partial f}{\partial y} \frac{dy}{dx}$$
Now I substitute in what is known and can simplify to reach the answer,
$$0=2x+2y\frac{dy}{dx} \ \ \ \ \ (*)$$
This leads to the correct answer, but I'm certain that my reasoning must be wrong. In $(*)$, I asserted that $\frac{dr^2}{dx}=0$, which is obvious. However, does this not also mean that the partial derivatives must be 0 instead of $2x$,$2y$? It does not make any intuitive sense to me how the partial derivative of a constant can be non-zero, and I am certain this cannot be the case.
 A: I think it is a confusion on understanding the Chain Rule. Your notation $\frac{df}{dx}$ does not even make sense. 
By the condition $x^2+y^2=r^2$, we know that $y$ is implicitly determined by $x$, say $y=y(x)$. If we define $h(x)=(x,y(x))$, then the composition $g(x)=f \circ h(x) = f(x,y(x))$ is then suitable to use the Chain Rule. We have: 
$$
\frac{dg}{dx}=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \cdot (1, \frac{dy}{dx})^T=2x+2y \frac{dy}{dx}.
$$
Since $g(x)=f(x,y(x))=x^2+(y(x))^2$, and $y(x)$ has the property that $x^2+(y(x))^2=r^2$, we have $g(x)=r^2$. And then $\frac{dg}{dx}=0$ by common sense, hence 
$$
2x+2y \frac{dy}{dx}=0
$$
is correct.
A: Let $y:\mathbb{R} \to \mathbb{R}$ be defined implicitly by $G(x,y(x)) = k$ for some $k \in \mathbb{R}$ with $G:\mathbb{R}^2 \to \mathbb{R}$. If $y$ and $G$ are differentiable and $\frac{\partial G}{\partial y} \neq 0$, then we can prove the implicitit differentiation formula:
$$
\frac{dy}{dx} = - \frac{\frac{\partial G}{\partial x}}{\frac{\partial G}{\partial y}}
$$
Proof: Define $H(x) = (G\circ F)(x)$ where $F:\mathbb{R} \to \mathbb{R}$ is defined by $F(x) = (x,y(x))$. Note that $H(x) = k$ so $H'(x) = \mathbf{D}H(x) = 0$ for all $x\ in \mathbb{R}$. But $H$ is also a composition of functions, so by the chain rule (for maps $\mathbb{R}^n \to \mathbb{R}^m)$ we have
\begin{align*}0 & = \mathbf{D}H(x)\\\\
& = \mathbf{D}G(H(x)) \cdot \mathbf{D}F(x)\\\
& = \begin{bmatrix}\frac{\partial G}{\partial x} & \frac{\partial G}{\partial y}\end{bmatrix} \cdot \begin{bmatrix}1 \\ \frac{dy}{dx}\\
\end{bmatrix}\\
& = \frac{\partial G}{\partial x} \cdot 1 + \frac{\partial G}{\partial y} \cdot \frac{dy}{dx}.
\end{align*}
Since $\frac{\partial G}{\partial y} \neq 0$, we can solve for $\frac{dy}{dx}$. This proves the claim.
I think the confusion here may be that you are assigning (in the above notation) $G(x,y) = r^2$, rather than $G(x,y) = x^2 + y^2$.  
