Solution
= Solution
The function is <smooth>, $f(0,0)=0$, and
$$
f_x=ye^x+e^y\cos x,\qquad f_y=e^x+e^y\sin x.
$$
At the origin both <partial derivatives> equal one. In particular, $f_y(0,0)=1\neq0$, so the <implicit function theorem> supplies a <neighbourhood> in which
$$
\boxed{C=\{(x,h(x))\},\qquad h\in C^1,\quad h(0)=0.}
$$
In fact $h$ is <smooth> because $f$ is <smooth>.