I'm solving the following ODE:
$$(1 + y^2) = (\arctan(y) - x)\frac{dy}{dx},$$
with $y(x)$ being the unknown function. The Solutions sheet recommends substituting $u(x) = \arctan(y(x))$, so that $\frac{du}{dx} = \frac{1}{y^2 + 1}\frac{dy}{dx}$. After making the substitution, the sheet reports the equation becoming
$$1 = (\arctan(y) - x)\frac{1}{y^2 + 1}\frac{dy}{dx}.$$
Why does $1 + y^2$ become $1$ after the substitution? Shouldn't it be $1 + \tan^2(u) = \sec^2(u)$?
You haven't done the substitution yet. The equation $$ 1 = (\arctan(y) - x) \frac{1}{y^2 + 1} \frac{dy}{dx} $$ is obtained by dividing the given ODE by the quantity $(y^2 + 1)$. Now, it becomes $$ 1 = (u - x) \frac{du}{dx}, $$ but this is still not in a familiar form. Notice that if you define the shifted variable $v = u - x$, so $u = v + x$, then $$ \dfrac{du}{dx} = \dfrac{dv}{dx} + 1, $$ so $$ 1 = v \biggl( \frac{dv}{dx} + 1 \biggr) $$ or $$ v\, \frac{dv}{dx} = 1 - v $$ which is separable. I trust that you can solve this equation for $v$. Then, you have to substitute back $v \mapsto u \mapsto x$.
Of course, this didn't have to happen in two steps since $$ v = u - x = \arctan y - x $$ could be a single substitution $v = \arctan y - x$, but that's hindsight. The path you take to solving the problem is not necessarily the most efficient, "cleanest" solution, but it helps build your intuition for next time.