$\displaystyle z = sin (u) cos (v); u=3x^2, v=x-3y$

$\displaystyle \left(\begin{array}{cc}cos(u)cos(v)&-sin(v)sin(u)\end{array}\right) \left( \begin{array}{cc} 6x&0\\1&-3\end{array}\right)$$\displaystyle =\left(\begin{array}{cc}6xcos(3x^2)cos(x-3y)-sin(x-3y)sin(3x^2)\\0-3sin(x-3y)sin(3x^2)\end{array}\right)$

I have a feeling this is incorrect