This question actually came up as I was reading an example in my differential equations book (Boyce & Diprima):
Solve: $2x+y^2+2xyy'=0$
Define $\psi(x,y)=x^2+xy^2$
Then $$\frac{\partial \psi}{\partial x}=2x+y^2, \frac{\partial \psi}{\partial y}=2xy$$
The differential equation is then $$\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx}=0$$
Here's what I don't understand:
$y$ is a function of $x$ - assume it's one-to-one. $\frac{\partial \psi}{\partial x}$, by definition, is computed for $y$ held constant. If $y$ is a function of $x$, if $y$ is held constant, $x$ must be constant too - what then is the meaning of $\frac{\partial \psi}{\partial x}$?
$\phi$ is actually defined on $\mathbb R^2$; the author could have written $$ \phi: \mathbb R^2 \to \mathbb R: (a, b) \mapsto a^2 + a b^2 $$
The problem is that our notation for derivatives depends on the choice of "dummy variables" that we use in denoting a function (sigh...on the good side, it often makes the chain rule work out in really nifty ways).
If, instead of $\partial \phi/\partial x$, you'd accept the notation $\partial_1 \phi$ to indicate "partial derivative with respect to the first argument", then you have
$$ \partial_1 \phi(a, b) = 2a + b^2 $$ or $$ \partial_1 \phi(x, y) = 2x + y^2 $$ or even $$ \partial_1 \phi(x, y(x)) = 2x + y(x)^2 $$ where $y$ is some function of the single variable $x$.