<s>[INST] Solving $-y' +y^2 = 1$ I must solve the ODE $-y' + y^2 = 1$. Rewriting this I can get: $$y' - y^2 + 1 = 0$$ However, because of the nonlinear term $y^2$ I do not know how to solve this. We cannot use the characteristic equation I think  and I cannot see that this is seperable.
Can someone please lend a hand?

I get the inverse of what WolframAlpha says the answer is. I get $y = \frac{e^{2x+2c} + 1 }{1 - e^{2x + 2c}}$.
From $\frac{dy}{dx} = y^2 - 1$ I write $\frac{dy}{y^2 - 1} = dx$ and then integrate: $$\int \frac{1}{y^2 - 1} dy = \int 1 dx = x + c_1$$
The integral on the left I decompose into partial fractions: $$\frac{1}{y^2 -1} = \frac{1}{(y-1)(y+1)} = \frac{A}{y-1} + \frac{B}{y+1}$$ and I get $A = \frac{1}{2}$ and $B = -\frac{1}{2}$
So then I have $\frac{1}{2}\int \frac{1}{y-1} - \frac{1}{2} \int \frac{1}{y+1} = x + c_1$ which is $$\frac{1}{2}ln(y-1)-\frac{1}{2}ln(y+1) = x + c$$ This simplifies to $$\frac{1}{2}ln(\frac{y-1}{y+1}) = x+c$$ Raising both sides by $e$: $$(\frac{y-1}{y+1})^{\frac{1}{2}} = e^{x+c}$$
Now squaring... $$\frac{y-1}{y+1} = e^{2x+2c}$$
Then: $$y - 1 = e^{2x+2c}y + e^{2x+2c} \implies y(1-e^{2x+2c}) = e^{2x+2c}+1$$
and finally $$y = \frac{e^{2x+2c} + 1}{1 - e^{2x+2c}}$$
 [/INST] Try $\displaystyle y' = \frac{dy} {dx} = y^2 - 1$ which is immediately separable and solvable by partial fraction decomposition. A hyperbolic trigonometric substitution will likely work as well, it's less elementary though.
Adding on, there is a very important subtlety here. The integral of $\displaystyle \frac 1x$ is not simply $\displaystyle \ln x + c$, it is actually $\displaystyle \ln |x| + c$. That absolute value sign is actually critical. It means that for the domain $\displaystyle x<0$, you should take $\displaystyle \ln (-x) + c_1$, while for $\displaystyle x>0$, you should take $\displaystyle \ln x + c_2$.
After integration, you wrote: $\displaystyle \frac 12\ln(y-1) - \frac 12\ln(y+1) = x + c$
This is imprecise (and not correct without domain restriction). For the general solution, consider three cases (note that $\displaystyle y$ cannot be equal to either $\displaystyle -1$ or $\displaystyle 1$):
Case 1: $\displaystyle y<-1$:
In this case, integration yields $\displaystyle \frac 12\ln(-(y-1)) - \frac 12\ln(-(y+1)) = x + c$  which, after you do the remaining working, will get you the exact solution you got, namely $\displaystyle y = \frac{1+e^{2(x+c)}}{1- e^{2(x+c)}}$
Case 2: $\displaystyle -1<y<1$:
In this case, integration yields $\displaystyle \frac 12\ln(-(y-1)) - \frac 12\ln(+(y+1)) = x + c$
Working through this would get you the same form as the Wolfram Alpha solution:
$\displaystyle y = \frac{1-e^{2(x+c)}}{1+e^{2(x+c)}}$
And finally,
Case 3: $\displaystyle y>1$ would get you, after integration:
$\displaystyle \frac 12\ln(+(y-1)) - \frac 12\ln(+(y+1)) = x + c$
Working through which you would get the exact same form as for case 1:
$\displaystyle y = \frac{1+e^{2(x+c)}}{1- e^{2(x+c)}}$
Splitting it up into cases gives you the full picture. You can visualise exactly what is going on by graphing.
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