= Solution
For $a\in U$, the <function> $f:U\to\mathbb R^n$ is <differentiable> at $a$ if there is a <linear map> $A:\mathbb R^m\to\mathbb R^n$ such that
$$
f(a+h)=f(a)+Ah+r(h),
\qquad
\frac{\lVert r(h)\rVert}{\lVert h\rVert}\longrightarrow0
\quad\hbox{as }h\longrightarrow0.
$$
The <linear map> is unique and is the <Fréchet derivative> $Df_a=A$. The <function> is <continuously differentiable> at $a$ if it is <differentiable> on a <neighbourhood> of $a$ and the map $x\mapsto Df_x$, with values in the space of <linear maps> equipped with the <operator norm>, is <continuous> at $a$.
If $T:\mathbb R^m\to\mathbb R^n$ is linear, then
$$
T(a+h)-T(a)-T(h)=0.
$$
Consequently $DT_a=T$ at every $a$. The <derivative> is a <constant function> of $a$, hence is continuous, so every <linear map> is continuously <differentiable> everywhere.
The <mean value inequality> says that if the <line segment> $[x,y]$ lies in $U$ and
$$
\lVert Df_z\rVert\leq M\qquad(z\in[x,y]),
$$
then
$$
\lVert f(y)-f(x)\rVert\leq M\lVert y-x\rVert.
$$
To prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\lVert v\rVert$ and apply the one-dimensional <mean value theorem> to the <real-valued function>
$$
\phi(t)=e\mathbin{\cdot}f\bigl(x+t(y-x)),
\qquad 0\leq t\leq1.
$$
For some $c\in(0,1)$, the <chain rule> gives
$$
\begin{aligned}
\lVert f(y)-f(x)\rVert
&=\phi(1)-\phi(0)\\
&=e\mathbin{\cdot}Df_{x+c(y-x)}(y-x)\\
&\leq \lVert Df_{x+c(y-x)}\rVert\lVert y-x\rVert
\leq M\lVert y-x\rVert,
\end{aligned}
$$
as required.
Now suppose that $U$ is <open> and <connected> and that $Df_a=0$ for every $a\in U$. Every point has an <open ball> contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is <locally constant>. Fix $a_0\in U$. The <level set>
$$
E=\{x\in U:f(x)=f(a_0)\}
$$
is nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the <zero derivative on a connected open set> result: $f$ is constant.
The <inverse function theorem> states that if $f:U\to\mathbb R^m$ is continuously <differentiable>, $a\in U$, and $Df_a$ is an <invertible linear map>, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f|_V:V\to W$ is a <bijection> whose inverse is continuously <differentiable>.
For the curve in the question, define the continuously <differentiable function>
$$
F(x,y)=x^2+y+\cos(xy)-1.
$$
Then
$$
F(0,0)=0,
\qquad
\frac{\partial F}{\partial y}(x,y)=1-x\sin(xy),
\qquad
\frac{\partial F}{\partial y}(0,0)=1\neq0.
$$
The <implicit function theorem>, which follows from the <inverse function theorem> applied to $(x,y)\mapsto(x,F(x,y))$, therefore gives an <open interval> $I$ containing $0$, an open <neighbourhood> $U_0$ of $(0,0)$, and a <continuously differentiable>, hence <continuous>, <function> $g:I\to\mathbb R$ such that
$$
\boxed{U_0\cap C
=\{(x,y)\in\mathbb R^2:x\in I,\ y=g(x)\}.}
$$
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