Is there a clear, universal test for a separable diferential equation? I understand that an equation of the form:
$$\frac{dy}{dx} = f(x, y)$$
is separable, if $f(x, y)$ can be rewritten as $g(x)\cdot h(y)$. 
But is there a way to test if the equation can be separated without having to guess until you find a valid separation? 
If there exists no universal method, is there a way to prove that a certain equation of a specific form cannot be separated, for example,$$ y' = \frac{x+y}{x}? $$
 A: Let $f(x,y) = g(x)\cdot h(y)$ and ignore any dependence of $y$ on $x$.  (I.e., do not view the following steps as implicit differentiation).
Let $G = \frac{\partial}{\partial x} f(x,y)$ and $H = \frac{\partial}{\partial y} f(x,y)$.
First, we should verify that $x$ and $y$ actually participate in the problem.  If $G = 0$, $x$ does not participate in the problem, so you have $y' = f(y)$, an autonomous differential equation.  If $H = 0$, $y$ does not participate in the problem, so you have $y' = f(x)$, and you should just integrate.  If both are zero, you have $y' = f$ which is just some constant, so your solution is a family of lines with slope $f$.
Notice, if your equation is separable, $G = g'(x) \cdot h(y) + g(x) \cdot 0 = g'(x) h(y)$.  So $\frac{G}{f} = \frac{g'}{g}$ has no dependence on $y$ at all.  You can verify this via $\frac{\partial}{\partial y} \frac{G}{f} = 0$.  Similarly, if your equation is separable, $\frac{\partial}{\partial x} \frac{H}{f} = 0$.  But you want to go the other way.
Suppose $\frac{\partial}{\partial y} \frac{G}{f} = \frac{\partial}{\partial x} \frac{H}{f} = 0$.  Integrating the first with respect to $y$, we get $\frac{G}{f} = c_1 + u(x)$ for some function of integration $u$.  (If you have never seen this before, this is equivalent to a constant of integration.  Notice if we differentiate with respect to $y$, $u(x)$ is sent to zero, in exactly the same way a normal, single variable constant of integration is.)  Integrating the second with respect to $x$, we get $\frac{H}{f} = k_1 + v(y)$.  So we have \begin{align*}
G &= \frac{\partial}{\partial x} f(x,y) = (c_1 + u(x))f  \\
H &= \frac{\partial}{\partial y} f(x,y) = (k_1 + v(y))f
\end{align*}
You should recognize these; their solutions are \begin{align*}
f(x,y) &= \mathrm{e}^{c_0 + c_1 x + \int u(x) \,\mathrm{d}x + S(y)} = \mathrm{e}^{c_0 + c_1 x + \int u(x) \,\mathrm{d}x} \mathrm{e}^{S(y)}  \\
f(x,y) &= \mathrm{e}^{k_0 + k_1 y + \int v(y) \,\mathrm{d}y + T(x)} = \mathrm{e}^{k_0 + k_1 y + \int v(y) \,\mathrm{d}y} \mathrm{e}^{T(x)}  \text{.}
\end{align*}
These can only both be true if (up to a multiplicative constant which we can partition arbitrarily between the factors) \begin{align*}
S(y) &= k_0 + k_1 y + \int v(y) \,\mathrm{d}y  \\
T(x) &= c_0 + c_1 x + \int u(x) \,\mathrm{d}x  \text{.}
\end{align*}
That is, we have forced 
$$ f(x,y) = \mathrm{e}^{S(y)} \cdot \mathrm{e}^{T(x)}  \text{,}  $$
to be separable.
