Differentiate $\,y = 9x^2 \sin x \tan x:$ Did I Solve This Correctly? I'm posting my initial work up to this point. 
Criticism welcomed!


Using the formula $(fgh)' = f'gh+fg'h + fgh'$, differentiate$$y = 9x^2\sin x \tan x$$

$$\begin{align} y' &= 9\frac d{dx}(x^2)\sin x \tan x + 9x^2 \frac d{dx} (\sin x) \tan x + 9x^2\sin x \frac d{dx}(\tan x)\\ \\
& = 9(2x\sin x \tan x + 9x^2-\cos x \tan x + 9x^2\sin x \sec^2x\\ \\
&=9\Big(2x\sin x \tan x + x^2 -\cos x \tan x + x^2\sin x \sec^2 x\Big)
\end{align}$$

Have I done it correctly up and until this point?

You have a parenthesis error
Line 3 should read
$$
9(2x)\sin x \tan x +9x^2(-\cos x)\tan x + 9x^2 \sin x \sec^2 x
$$
You can actually factor out $9x\sin x$ because $\cos x \tan x = \sin x$.
Continuing
$$
9x \sin x (2\tan x - x + x \sec^2 x) \\
9x \sin x (2\tan x +x (\sec^2 x - 1) )\\
$$
Then you can use some trig identities to simplify further.
