Question: Let $f \colon \mathbb{R}^2 \to \mathbb{R}$ be a continuous function so that $\frac{\partial f}{\partial y}$ exists and is continuous on $\mathbb{R}^2$. Let $(x_0, y_0) \in \mathbb{R}^2 $ be a point where $f(x_0, y_0) = 0$ and $\frac{\partial f}{\partial y}(x_0, y_0) > 0$. Show that there are $u,v \in \mathbb{R}, u,v > 0$ such that $|x - x_0| < u$ and there is a unique $y$ with $|y - y_0| < v$ so that $f(x,y) = 0$.
My attempt: If $\frac{\partial f}{\partial x}$ exists and continuous as well, then the result holds for any $(x,y)$ in a neighbourhood about $(0,0)$ via the direct application of implicit function theorem. However, is there any way I can tackle this problem using inverse function theorem or etc to obtain the unique $y$ without knowledge of the existence of $\frac{\partial f}{\partial x}$ and also continuity of $\frac{\partial f}{\partial x}$? Any hints are appreciated.