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,$$ y' = \frac{x+y}{x}? $$
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.