Why can't I manipulate the entire equation?
You can. The analytical method for proving an identity consists of starting with the
identity you want to prove, in the present case
$$
\begin{equation}
\frac{\sin \theta -\sin ^{3}\theta }{\cos ^{2}\theta }=\sin \theta,\qquad \cos \theta \neq 0
\tag{1}
\end{equation}
$$
and establish a sequence of identities so that each one is a consequence of
the next one. For the identity $(1)$ to be true is enough that the following
holds
$$
\begin{equation}
\sin \theta -\sin ^{3}\theta =\sin \theta \cos ^{2}\theta \tag{2}
\end{equation}
$$
or this equivalent one
$$
\begin{equation}
\sin \theta \left( 1-\sin ^{2}\theta \right) =\sin \theta \cos ^{2}\theta
\tag{3}
\end{equation}
$$
or finally this last one
$$
\begin{equation}
\sin \theta \cos ^{2}\theta =\sin \theta \cos ^{2}\theta \tag{4}
\end{equation}
$$
Since $(4)$ is true so is $(1)$.
The book indicated below illustrates this method with the following identity
$$
\frac{1+\sin a}{\cos a}=\frac{\cos a}{1-\sin a}\qquad a\neq (2k+1)\frac{\pi
}{2}
$$
It is enough that the following holds
$$
(1+\sin a)(1-\sin a)=\cos a\cos a
$$
or
$$
1-\sin ^{2}a=\cos ^{2}a,
$$
which is true if
$$
1=\cos ^{2}a+\sin ^{2}a
$$
is true. Since this was proven to be true, all the previous indentities
hold, and so does the first identity.
Reference: J. Calado, Compêndio de Trigonometria, Empresa Literária
Fluminense, Lisbon, pp. 90-91, 1967.
Best Answer
Note that the numerator of your derivative is $-2(\cos^2x+\sin^2x)$, which does indeed simplify to $-2$. It seems you mistook the numerator for $-2(\cos^2x-\sin^2x)$ (which would simplify to $-2\cos 2x$), which is an understandable typo-level mistake.
For your general question, there's no hard and fast rule for when you will need/want to use particular identities; it all depends on the goal of your computations and the specific opportunities you see.