If $x=t^2\sin3t$ and $y=t^2\cos3t$, find $\frac{dy}{dx}$ in terms of $t$ If $x=t^2\sin3t$ and $y=t^2\cos3t$, find $\frac{dy}{dx}$ in terms of $t$. This is how I tried solving it:
$$
\frac{dx}{dt} = 2t\sin3t + 3t^2\cos3t \\
\frac{dy}{dt} = 2t\cos3t - 3t^2\sin3t \\
\frac{dy}{dx} = \frac{2t\cos3t - 3t^2\sin3t}{2t\sin3t + 3t^2\cos3t} 
$$
But the answer listed is: 
$$
\frac{2-3t\tan3t}{2\tan3t+3t}
$$
Is my answer incorrect, or can I simplify it even more?

What you did is entirely correct. To show that it is equivalent to the second solution, divide numerator and denominator by $t \cos 3t$:
$$\frac{dy}{dx} = \frac{2t\cos3t - 3t^2\sin3t}{2t\sin3t + 3t^2\cos3t}\cdot \frac{\frac 1{t\cos 3t}}{\frac{1}{t\cos 3t}} = \frac{2-3t\tan 3t}{t\tan 3t+ 3t}$$
