In our problem sheet on Real Analysis, there is a problem that seems to be unprecise (perhaps it's just me - I'm sorry if that's the case.). I'm asking you to please only answer on what the notation should say (and not spoil anything on the problem).
Let $f:\mathbb{R}^2 \to \mathbb{R}$ be differentiable. Show that a differentiable function $y: \mathbb{R} \to \mathbb{R}$ solves the differential equation $$ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}y' = 0$$ if and only if there exists a $c \in \mathbb{R}$ with $$ f(x,y(x)) = c$$ for all $x \in \mathbb{R}$.
So should that differential equation stand for $$ \frac{\partial f}{\partial x_1} (x_1,x_2) + \frac{\partial f}{\partial x_2}(x_1,x_2)y(x_1) = 0 $$ for all $(x_1,x_2) \in \mathbb{R}^2$ or what exactly should it say? To which variables do my instructors intend to refer these functions?
It's a bit awkward to not touch on the content of the question but just the notation ...
That differential equation is explicitly writing out the total derivative $\displaystyle \frac{ \mathrm{d}f }{ \mathrm{d}x_1 }$ when $f$ is expressed as a function of both $x_1$ and $x_2$ but while $x_2$ is "secretly" a function of $x_1$. $$\frac{\partial f}{\partial x_1} (x_1,x_2) + \frac{\partial f}{\partial x_2}(x_1,x_2) \frac{ \mathrm{d} x_2(x_1) }{ \mathrm{d} x_1} = 0$$ for all points along the trajectory $\left(x_1, x_2(x_1) \right)$ for all $x_1 \in \mathbb{R}$.
Seriously, this is just replacing the notation $(x,y)$ in the given question with your $(x_1, x_2)$.