How to prove such a relationship? I have a function $S(x,y)$ which satisfies the following PDE
$$\frac{\partial S(x,y)}{\partial y}=-H\left(x,\frac{\partial S(x,y)}{\partial x}\right)$$
where the known function $H(x,z)=x^2+xz+z^2+\frac{1}{x-\alpha}+\frac{1}{\beta-x}$. Here $\alpha<x<\beta$.

Can we prove that when $x\to\alpha$ or $x\to\beta$, $\dfrac{\partial S(x,y)}{\partial x}\to\infty$?

Thanks in advance!
 A: Assume, without loss of generality, that $x \to \alpha$ (keep in mind that $x \to \alpha+0$). Then there are three possible limits for $H(x,S'_x)$ as $x \to \alpha$:
1) $H(x, S'_x) \to +\infty$;
2) $H(x, S'_x) \to C$, where $C$ is some constant;
3) $H(x, S'_x) \to -\infty$.
It is not hard to show, that 2) and 3) are impossible. Indeed, assume first that $|S'_x| < C_1 < +\infty$ for all $x \in [\alpha, \beta)$. Then, of course, $H(x, S'_x) \to +\infty$ (just because all the terms, except $\frac{1}{x-\alpha}$, are bounded).
If now $S'_x \to \pm \infty$ (the sign of $\infty$ is irrelevant), then
$$
x S'_x + (S'_x)^2 = S'_x (x + S'_x) \to +\infty,
$$
and also $\frac{1}{x-\alpha} \to +\infty$. Hence, we get that only 1) is possible.
Now, we assume that $S(x,y) = F(x) G(y)$.
Since 
$$
H(x, S'_x) \equiv H(x, F'(x) G(y)) \to +\infty,
$$
we get (from the equation) that
$$
S'_y = F(x) G'(y) \to -\infty \quad \text{as } x \to \alpha.
$$
However, this implies that $F(x) \to -\infty$, since $G'(y)$ is a fixed constant for each fixed $y$.
But, this implies that $F'(x) \to -\infty$, since $F(x)$ is continuously differentiable on $(\alpha, \beta)$.
Indeed, if we suppose that $|F'(x)| < C_1 < +\infty$ for all $x \in [\alpha, \beta)$, then by, for example, the mean value theorem we get that $|F(\alpha)| < C_2 < +\infty$, which is a contradiction.
Therefore, $F'(x) \to -\infty$, and, consequently, $S'_x = F'(x) G(y) \to -\infty$.
