Let $F(x,y)$ be a function and let $P=(x_{0},y_{0})$ be a point on the curve $F(x,y)=0$. The implicit function theorem tells us that if $F(x,y)$ satisfies the following two conditions:
(1) $F(x,y)$ is continuously differentiable on a neighbourhood of $P$;
(2) $\dfrac{\partial F}{\partial y}(x_0,y_0)\neq0$.
Then for the curve $F(x,y)=0$ around $(x_0, y_0)$, we can write $y=f(x)$, where $f(x)$ is a real function.
My question is, if we change the first condition to: $F(x,y)$ is just differentiable at $P$, the partial derivative of $F$ maybe not continuous, can we determine an implicit function in the neighborhood of point $P$(here we don't care about whether the implicit function is derivable)?
I rewrite the answers of Ted Shifrin and GReyes. If $f(x)$ and $g(x)$ are differentiable functions, then $F(x,y)=f(x)+g(y)$ is a differentiable function of two variables and we have that $$F_x(x,y)=f'(x),\qquad F_y(x,y)=g'(y).$$ Thus, if one of $f'(x)$ and $g'(y)$ is not continuous, then $F(x,y)$ is a differentiable function but not a $C^1$ function. Now let $$f(x)=\sin x,\quad g(y)=\begin{cases}y/2+y^2\sin(1/y),&y\neq 0\\ 0,&y=0\end{cases}$$ and $$F(x,y)=f(x)+g(y),$$ then $$F_y(x,y)=\begin{cases}1/2+2y\sin(1/y)-\cos(1/y),&y\neq 0\\ 1/2,&y=0\end{cases}.$$ Special, $F_y(0,0)=1/2\neq0$. But in any neighbour $U$ of point $(0,0)$, $y$ is not a function of $x$. In fact, for any neighbour $(-\delta,\delta)$ of $x=0$ and every point $x_0\in(-\delta,\delta)$, there are infinitely many $y_n$ which trend to zero such that $F(x_0,y_n)=0$.