You have a transformation group acting on the plane, say $SO(2)$, and some curve in the plane, say $(x,u(x))$. The image of the curve is then $(X(x),U(x))=(x\cos\theta-u\sin\theta,x\sin\theta+u\cos\theta)$. THe derivative of the new curve can be found by $$\frac{dU}{dX} = \frac{dU/dx}{dX/dx} = \frac{\sin\theta+u'\cos\theta}{\cos\theta-u'\sin\theta}$$
I'm stuck on what to do with a surface in three-dimensions, say $(x,y,u(x,y))$ begin transformed to $(X(x,y),Y(x,y),U(x,y))$. How do I find the partial derivatives $$\frac{\partial U}{\partial X} \ \ \mbox{and} \ \ \frac{\partial U}{\partial Y} \ \ ?$$
EDIT
I seem to remember two equations with sums of products of derivatives coming from the chain rule. Then they're put into a matrix equation and inverted. Not sure if that's a red herring.
If we want $U_X$ and $U_Y$ then we are thinking of $U$ as a function of $X$ and $Y$, but $X$ and $Y$ are themselves functions of $x$ and $y$, meaning we have $U(X(x,y),Y(x,y))$.
Differentiating with respect to $x$ and $y$ in turn gives \begin{eqnarray*} \frac{\partial U}{\partial x} &=& \frac{\partial U}{\partial X}\frac{\partial X}{\partial x}+\frac{\partial U}{\partial Y}\frac{\partial Y}{\partial x} \\ \\ \frac{\partial U}{\partial y} &=& \frac{\partial U}{\partial X}\frac{\partial X}{\partial y}+\frac{\partial U}{\partial Y}\frac{\partial Y}{\partial y} \end{eqnarray*}
In terms of a matrix equation:
$$\left(\begin{array}{c} U_x \\ U_y \end{array}\right) = \left(\begin{array}{cc} X_x & Y_x \\ X_y & Y_y \end{array}\right) \left(\begin{array}{c} U_X \\ U_Y \end{array}\right)$$
Inverting this gives $$\left(\begin{array}{c} U_X \\ U_Y \end{array}\right)=\frac{1}{X_xY_y-X_yY_x} \left(\begin{array}{cc} Y_y & -Y_x \\ -X_y & X_x \end{array}\right)\left(\begin{array}{c} U_x \\ U_y \end{array}\right)$$