Finding approximation to stable manifold of saddle point I am stuck on the following exercise from Strogatz' book on dynamical systems (exercise 6.1.14).

Consider the system $\dot{x} = x+e^{-y}, \dot{y} = -y$. This system
  has a single fixed point, $(-1, 0)$. This is a saddle point. The
  unstable manifold is $y=0$, but the stable manifold is some non-linear
  curve. Let $(x, y)$ be a point on the stable manifold close to $(-1,
> 0)$ and define $u = x + 1$. Write the stable manifold as $y=a_1u +a_2u^2 + O(u^3)$. To determine the coefficients, derive two
  expressions for $dy/du$ and equate them.

I have, as a first try for an equation, simply differentiated $y$ wrt $u$: $\frac{dy}{du} = a_1 + 2a_2u + O(u^2)$, where I'm a bit uncertain about the $O(u^2)$, but I suspect we can leave that out anyway, since we're approximating. I'm not sure how we'd find a second equation.
I've found a post that answers the same question, but it ends up with a line, rather than the non-linear curve that is pictured in the book:

Furthermore, I don't see why they have $\frac{dy}{du} = \frac{\dot{y}}{\dot{u}}$ or how they calculated the Taylor approximation for $\dot{u}$.
Basically, I'm a bit lost, could someone perhaps give me some hints?
 A: Remember that
$$
\frac{{\rm d}y}{{\rm d}u} = \frac{{\rm d}y / {\rm d}t}{{\rm d}u / {\rm d}t} \tag{1}
$$
If you replace 
$$
\frac{{\rm d}y}{{\rm d}t} = -y \tag{2}
$$
and 
$$
\frac{{\rm d}u}{{\rm d}t} = \frac{{\rm d}x}{{\rm d}t} = u - 1 + \left(1 - y + \frac{y^2}{2} - \cdots \right) \tag{3}
$$
In (1) you will get
$$
\frac{{\rm d}y}{{\rm d}u} = -\frac{y}{u - y + y^2/2 - y^3/6 + \cdots} \tag{4}
$$
Now you replace your expression for $y$: $y(u) = a_1u + a_2 u^2 + \cdots$. I'm going to take another path to the solution you just linked, and expand the result to a higher order, you do this by replacing the expression for $y$ in (4) and Taylor expanding it around $u = 0$, the result is
\begin{eqnarray}
\frac{{\rm d}y}{{\rm d}u} &=& \frac{a_1}{a_1 - 1} + \frac{a_1^3 - 2 a_2 }{2(a_1 - 1)^2}u + \frac{2a_1^4 + a_1^5 - 18a_1^2 a_2 + 12 a_2^2 + 12 a_3 - 12 a_1 a_3}{12(a_1 - 1)^3}u^3 + \cdots \\
&=& a_1 + 2a_2 u + 3a_3 u^3 + \cdots
\end{eqnarray}
You solve for coefficients, and you should get
$$
a_1 = 2,~~ a_2 = 4/3, ~~a_3 = 10/9
$$
The black line below is the result

