\noindent{\bf Supplementary problem 1: }
We have 
$$
f'(x,y) = \left(\begin{array}{cc}
\cos x & \sin y \\
e^x & e^y
\end{array}\right)
$$
which clearly varies continuously with $(x,y)$.  So $f$ is a $C^1$ function
on all of $\R^2$.
\begin{itemize}
\item We check
$$
\det f'(0,0) = \det \left(\begin{array}{cc}
1 & 0 \\
1 & 1
\end{array}\right)
= 1.
$$
In particular $f'(0,0)$ is invertible, and the inverse function theorem
implies that $f$ is locally invertible at $(0,0)$.  The inverse function
has linear approximation at $(-1,2)$ given by
\begin{eqnarray*}
f^{-1}(-1,2) + (f^{-1})'(-1,2)
\left(\begin{array}{c} x+1 \\ y-2 \end{array}\right)
& = &   
\left(\begin{array}{c} 0 \\ 0 \end{array}\right)
+ 
f'(0,0)^{-1}
\left(\begin{array}{c} x+1 \\ y-2 \end{array}\right)
\\
& = & 
\left(\begin{array}{cc}
1 & 0 \\
-1 & 1
\end{array}\right)
\left(\begin{array}{c} x+1 \\ y-2 \end{array}\right) \\
& = &
\left(\begin{array}{c} x+1 \\ y-x-3 \end{array}\right)
\end{eqnarray*}

\item Let $(x_0,y_0) = (0,0)$ be a first guess at the desired point $(x,y)$,
and let us use Newton's method (and Mathematica!)
$$
(x_{n+1},y_{n+1}) = (x_n,y_n) - f'(x_n,y_n)^{-1}(f(x_n,y_n) - (-1.02,1.97))
$$
to improve our guesses til they settle down to three unchanging decimal
places. 

Thus,
\begin{eqnarray*}
\left(\begin{array}{c} x_1 \\ y_1 \end{array}\right)
 & = & \left(\begin{array}{c} 0 \\ 0 \end{array}\right) - 
\left(\begin{array}{cc}
1 & 0 \\
1 & 1
\end{array}\right)^{-1}\left(
\left(\begin{array}{c} -1 \\ 2 \end{array}\right)
-
\left(\begin{array}{c} -1.02 \\ 1.97 \end{array}\right)
\right) \\
& = &
-\left(\begin{array}{cc}
1 & 0 \\
-1 & 1
\end{array}\right)
\left(\begin{array}{c} .02 \\ .03 \end{array}\right)
= 
\left(\begin{array}{c} -.02 \\ -.01 \end{array}\right)
\end{eqnarray*}
And again
\begin{eqnarray*}
\left(\begin{array}{c} x_2 \\ y_2 \end{array}\right)
 & = & \left(\begin{array}{c} -.02 \\ -.01 \end{array}\right) - 
\left(\begin{array}{cc}
.9998 & -.0100 \\
.9802 & .9900
\end{array}\right)^{-1}\left(
\left(\begin{array}{c} -1.01995 \\  1.970255\end{array}\right)
-
\left(\begin{array}{c} -1.02 \\ 1.97 \end{array}\right)
\right) \\
& = &
\left(\begin{array}{c} -.0200533 \\  -.0101982 \end{array}\right)
\end{eqnarray*}
A further iteration of Newton's method shows that $(x_3,y_3)$ is
exactly the same as $(x_2,y_2)$ to the number of digits shown above,
so $(x_2,y_2)$ ought to be approximate $(x,y)$ accurately to at least 
5 (non-zero) digits.





\end{itemize}