Here, I will provide an answer which do not involve any use of Taylor
series. Also it uses L'Hospital's Rule a minimum of times, and in order to
prove some basic limits only which are the following
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6},\ \ \ \ \
\ \ \ \ and\ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{\sin ^{-1}x-x}{%
x^{3}}=\frac{1}{6} \\
\lim_{x\rightarrow 0}\frac{\sin x-x+\frac{x^{3}}{6}}{x^{5}} &=&\frac{1}{120},%
\ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{%
\sin ^{-1}x-x-\frac{x^{3}}{6}}{x^{5}}=\frac{3}{40}.
\end{eqnarray*}
First, note that the simple change of variable $\sin x=t$ shows that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{x\sin (\sin x)-\sin ^{2}(x)}{x^{6}}
&=&\lim_{x\rightarrow 0}\frac{x\sin (\sin x)-\sin ^{2}(x)}{\sin ^{6}x}\left(
\frac{\sin x}{x}\right) ^{6} \\
&=&\lim_{t\rightarrow 0}\frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}.
\end{eqnarray*}
Thus, actually, it is equivalent to answer the following equivalent question:
\begin{equation*}
\lim_{t\rightarrow 0}\frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}.
\end{equation*}
At present, some algebraic manipulations have to be used. But before, I will
explain how they are performed. If you want to write the number $56$ as a
function of the number $15$ you can write
\begin{equation*}
56=15\times 3+rest
\end{equation*}
and you find the $rest$ as
\begin{equation*}
rest=56-15\times 3=56-45=11.
\end{equation*}
The number $3$ was chosen (in $15\times 3$) so that the rest $11$ would be
smaller that $15.$ Let's go back to our expression
\begin{equation*}
\left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right)
\end{equation*}
we want to write it as a product of some expressions of the basic limits
cited above. Since the expression contains the product $\sin t\sin ^{-1}t$
and the denominator is $t^{6}$ so the natural choice would be this product $%
\left( \frac{\sin t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}%
\right) ,$ then we try to find the 'rest' as follows
\begin{equation*}
rest=\left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right) -\left( \frac{\sin
t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}\right)
\end{equation*}
which after easy computations one gets
\begin{eqnarray*}
rest &=&\left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right) -\left( \frac{%
\sin t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}\right) \\
&=&\frac{t\sin t+t\sin ^{-1}t-2t^{2}}{t^{6}}.
\end{eqnarray*}
Now we simplify by $t$ and try to write the result as a new function (maybe
product) of the expressions of basic limits. Since $\sin t$ and $\sin ^{-1}t$
are linear in the last expression, I mean there is no exponent on either of
them, so no product is permitted now, but we should add expressions
containing linear terms of $\sin t$ and $\sin ^{-1}t.$ The natural choice is
the following
\begin{equation*}
\frac{\sin t+\sin ^{-1}t-2t}{t^{5}}=\left( \frac{\sin t-t+\frac{t^{3}}{6}}{%
t^{5}}\right) +\left( \frac{\sin ^{-1}t-t-\frac{t^{3}}{6}}{t^{5}}\right) .
\end{equation*}
Now we resume the resulting computations as follows:
\begin{eqnarray*}
\left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right) &=&\left( \frac{\sin t-t%
}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}\right) +\frac{t\sin
t+t\sin ^{-1}t-2t^{2}}{t^{6}} \\
&=&\left( \frac{\sin t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}%
\right) +\frac{\sin t+\sin ^{-1}t-2t}{t^{5}} \\
&=&\left( \frac{\sin t-t}{t^{3}}\right) \left( \frac{\sin ^{-1}t-t}{t^{3}}%
\right) +\left( \frac{\sin t-t+\frac{t^{3}}{6}}{t^{5}}\right) +\left( \frac{%
\sin ^{-1}t-t-\frac{t^{3}}{6}}{t^{5}}\right)
\end{eqnarray*}
Without the explanations given above these decompositions look tricky or
very genus, as they fall down from the sky, but I hope I have provided
sufficient details to make them natural.
At the end, passing to the limit as $t$ tends to $0$ and using the basic limits gives immediately
\begin{equation*}
\lim_{t\rightarrow 0}\left( \frac{\sin t\sin ^{-1}t-t^{2}}{t^{6}}\right)
=\left( -\frac{1}{6}\right) \left( \frac{1}{6}\right) +\left( \frac{1}{120}%
\right) +\left( \frac{3}{40}\right) =\frac{1}{18}.\ \blacksquare
\end{equation*}
Now, permit me to bring to reader's attention that we have ONLY used BASIC
limits, and not the functions involved themselves. Also, if we remove the
extra explanations the answer is short.
Let $f$ and $g$ any two functions defined around $t=0$ such that
\begin{eqnarray*}
\lim_{t\rightarrow 0}\frac{f(t)-t}{t^{3}} &=&a,\ \ \ \ \ \ \ \ \ and\
\ \ \ \ \ \ \ \ \ \lim_{t\rightarrow 0}\frac{g(t)-t}{t^{3}}=b \\
\lim_{t\rightarrow 0}\frac{f(t)-t+\frac{t^{3}}{6}}{t^{5}} &=&c,\ \ \ \
\ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \lim_{t\rightarrow 0}\frac{g(t)-t-\frac{%
t^{3}}{6}}{t^{5}}=d.
\end{eqnarray*}
Then one can show that
\begin{equation*}
\lim_{t\rightarrow 0}\left( \frac{f(t)g(t)-t^{2}}{t^{6}}\right) =ab+c+d.
\end{equation*}
PROOF (which is short).
\begin{eqnarray*}
\left( \frac{\mathsf{f(t)g(t)}-t^{2}}{t^{6}}\right) &=&\left( \frac{\mathsf{%
f(t)}-t}{t^{3}}\right) \left( \frac{\mathsf{g(t)}-t}{t^{3}}\right) +\frac{t%
\mathsf{f(t)}+t\mathsf{g(t)}-2t^{2}}{t^{6}} \\
&=&\left( \frac{\mathsf{f(t)}-t}{t^{3}}\right) \left( \frac{\mathsf{g(t)}-t}{%
t^{3}}\right) +\frac{\mathsf{f(t)}+\mathsf{g(t)}-2t}{t^{5}} \\
&=&\left( \frac{\mathsf{f(t)}-t}{t^{3}}\right) \left( \frac{\mathsf{g(t)}-t}{%
t^{3}}\right) +\left( \frac{\mathsf{f(t)}-t+\frac{t^{3}}{6}}{t^{5}}\right)
+\left( \frac{\mathsf{g(t)}-t-\frac{t^{3}}{6}}{t^{5}}\right)
\end{eqnarray*}
Passing to the limit when $t$ tends to $0$ and using the BASIC LIMITS given in the hypothesis one gets
\begin{equation*}
\lim_{t\rightarrow 0}\left( \frac{\mathsf{f(t)g(t)}-t^{2}}{t^{6}}\right)
=ab+c+d.\ \blacksquare
\end{equation*}
