How to evaluate partial derivative of $f$ with respect to $x$ as a function of $y$ only? Hopefully this is just the notation confusing me, but how would I correctly evaluate the following function $g$ below?
$g(x,y) = \frac{\partial f(2y, y)}{\partial x}$ (or equivalently $g(x,y) = f_{x}(2y, y)$), where $f(x,y) = x + y$.
On one hand, $\frac{\partial f(x, y)}{\partial x} = 1$, so we could have $g(x,y) = 1$. On the other hand, $f(2y, y) = 3y$, suggesting $g(x,y) = \frac{\partial (3y)}{\partial x} =0.$
Edit:
To add some context, here is where this came from ($\phi$ below is a function of $x$ and $t$):
Define the mapping $T$ from $C^{\infty}_{0}(\mathbb{R}^2)$ to $\mathbb{R}$ by $T: \phi \mapsto \int_{0}^\infty \phi(t, ct)dt$, where $c$ is a real number. What is $\partial_x T$?
 A: I think the notation is meant to be understood as
$$g(x,y) \enspace = \enspace \frac{\partial f(x,y)}{\partial x} \; \Big|_{x = 2y} $$
So, your answer $g(x,y) = 1$ would be correct. However, I would rather consider the context carefully - hard to say what was really meant here since the whole notation seems a bit off, e.g. the LHS still holds the dependency for $x$, whilst the RHS doesn't. So, I'd probably write it as
$$ g(x = 2y, y ) \enspace = \enspace \frac{\partial f(x,y)}{\partial x} \; \Big|_{x = 2y} $$
$${}$$
EDIT:
With respect to your latest edit, I would understand your map as follows: The $x$-variable of $\phi(t,x)$ is a function of $t$, so one can rewrite it as $\phi \big( t, x(t) \big)$.
Therefore, your derivative can be solved as:
$$ \partial_x T \enspace = \enspace \partial_x T \big( \phi \big) \enspace = \enspace \partial_x T \big( \phi(t,x(t)) \big)$$
$$= \enspace \int_0^{\infty} \partial_x \phi \big( t, x(t) \big) \, dt \enspace = \enspace \int_0^{\infty} \frac{\partial \phi \big( t, x(t) \big)}{\partial x} \, dt$$
So, you would have to derive $\phi(t,ct)$ as if $ct$ was $x$ and then - AFTER deriving it, but still BEFORE integrating it - fill in $x = ct$. However, I am not entirely sure.
