Suppose that we have an invertible map $T(u,v)=(x,y)$. The Jacobian of $T$ is given by
$ \text{Jac}(T)= \left| {\begin{array}{cc} x_u & x_v \\ y_u & y_v \\ \end{array} } \right|$.
Now if we consider the inverse map of $T$, i.e. $T^{-1}(x,y)=(u,v)$, its Jacobian is given by
$ \text{Jac}(T^{-1})= \left| {\begin{array}{cc} u_x & u_y \\ v_x & v_y \\ \end{array} } \right|$.
Since $H=TT^{-1}=I$ is the mapping $H(u,v)=(u,v)$, we expect that
$ \text{Jac}(H)= \left| {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right|$.
I know that $1=\frac{du}{du}=\frac{dv}{dv}$ and that $0=\frac{du}{dv}=\frac{dv}{du}$ but I cannot seem to figure out how multiplying the two matrices above would give the identity matrix. Doesn't look like the right chain rule to me...any idea where I went wrong?
The multivariable chain rule is not normally stated in terms of strange things like $\frac{dv}{du}$. It is stated in terms of total derivatives, i.e., linear transformations that approximate the map at a given point. In particular, it says that if $T(u_0,v_0)=(x_0,y_0)$, then the derivative of the composition $T^{-1}\circ T$ at $(u_0,v_0)$ is the composition of the derivative of $T$ at $(u_0,v_0)$ and the derivative of $T^{-1}$ at $(x_0,y_0)$. The composition $T^{-1}\circ T$ is the identity map, so its derivative is also the identity map. Conclusion: the derivative of $T^{-1}$ at $(x_0,y_0)$ is the inverse of the derivative of $T$ at $(u_0,v_0)$. This can be expressed in matrices if one wishes.