How to prove that $y'=e^{xy}$ is not separable? Very new to this subject. I know that the definition for separable functions is that if it can be written as follows
$$
y'=g(x)h(y)
$$
then it is separable. There are many functions such as $y'=x+y$ or $y'=e^{xy}$ where you can see that it is not separable but how to actually prove that?
I tried to use the definition and write
$$
e^{xy}=g(x)h(y)
$$
but could not find a clear contradiction.
 A: We will suppose that, 
$$e^{xy} = g(x)h(y),$$
in hopes of reaching a contradiction. Now take the partial derivative with respect to $y$. 
$$ x e^{xy} = g(x) h'(y)$$
$$ x g(x)h(y) = g(x) h'(y)$$
Note that $g(x)$ cannot be zero for any value of $x$ because $e^{xy}$ is nonzero for all $x$ and $y$. This means we can divide both sides by $g(x)$ cancelling it from the equation. The same reasoning demonstrates that $h(y)$ is not zero for any $y$ and we can therefore also divide by $h(y)$. 
$$ x h(y) = h'(y)$$
$$ x = \frac{h'(y)}{h(y)}$$
Now on the right hand side we have a function of only $y$ and on the left hand side we have a function of only $x$ which is contradictory unless both sides equal a constant value; but the left hand side is not a constant value so we have arrived at a contradiction.
Therefore the initial assumption that $e^{xy}$ can be written as $g(x)h(y)$ is not true. 

Why is it a contradiction if we have a fucntion on only $y$ on the right and $x$ on the left?
$$ f(x) = g(y) $$
Well if both functions are equal to the same constant there is no contradiction. However in our case the left hand side wasn't a constant, in particular we had $f(x) = x$. 
$$ x = g(y) $$
Now lets plug in two ordered pairs for $(x,y)$. Lets choose $(1,2)$ and $(2,2)$. This would produce the following equations. 
$$ 1 = g(2) \qquad 2 = g(2)$$
which gives us two values for $g(2)$. However since $g$ is a function there is only one possible value for $g(2)$. This would require $1=2$ which is a contradiction because $1\neq 2$. 

Another way to see it is to take the partial derivative with respect to $x$ on both sides of the equation. 
$$ x = f(y)$$
$$ \frac{\partial x}{\partial x} = \frac{\partial f(y)}{\partial x}$$
$$1 = 0$$
This is a contradiction. 
A: In this case it is clear that $g(x)\ne0$ for all $x$. Consider two different values of $x$, say $a$ and $b$. Then, if $g$ and $h$ exist, we get:
$$h(y) = C_a e^{ay} = C_b e^{by}\,,$$
with $C_a=1/g(a)$ and $C_b=1/g(b)$. Taking $\ln$ in the previous equation we get:
$$ ay+\ln(C_a) = by+\ln(C_b)\,,$$
which is impossible if $a\ne b$ as required.
