The title was too short to sumarize my question. I'm currently trying to do the following exercise:
Let $f:U \rightarrow \mathbb{R}$ be continuous on an open $U \subset \mathbb{R}^2$, with partial derivative $\frac{\partial f}{\partial y}(a, b)>0$ at a point $(a, b) \in U$, with $f(a, b)=c$.
(1) Prove that there exists a rectangle $I \times J$, with sides parallel to the axis, contained in $U$ such that $f(x, y)>c$ when $(x, y)$ is on the upper base of the rectangle and $f(x, y)<c$ when $(x, y)$ is on the lower base.
(2) Show that there exists an interval $I$ centered at $a$, on which it is defined a function $\psi:I \rightarrow J$ such that $\psi(a)=b$ e, for every $x\in I$, $f(x, \psi(x))=c$. This function is not necessarily unique, but any such function is continuous on $x=a$.
(3) If $\frac{\partial f}{\partial y}>0$ at every point of $U$, show that $\psi$ is unique and is continuous on $I$.
Here's the progress I've made so far:
(1) Let $\lambda:\pi_2(U) \rightarrow U$ be defined by $\lambda(x)=f(a, x)$. Then $\lambda$ is continuous since so is $f$. Moreover, $\lambda$ is differentiable at $x=b$ with $\lambda'(b)=\frac{\partial f}{\partial y}(a, b)>0$. Then there exists an interval $J=(b-\delta, b+\delta)$ at which $\lambda$ is strictly increasing. That is, $f(a, y)>f(a, b)=c$ when $b<y<b+\delta$ and $f(a, y)<c$ when $b-\delta<y<b$. We know that $f(a, x)$ increases on $(b-\delta, b+\delta)$, so we may assume without of loss of generality that $f$ increases on $[b-\delta, b+\delta]$ (if it doesn't we just take a smaller $\delta$). So $f(a, b+\delta)>c$ and $f(a, b-\delta)<c$. By continuity of $f$, we can find $\epsilon_1, \epsilon_2>0$ such that $f(x, b+\delta)>c$ whenever $x\in (a-\epsilon_1, a+\epsilon_2)$ and $f(x, b-\delta)<c$ whenever $x\in (a-\epsilon_2, a+\epsilon_2)$. By letting $\epsilon< \min\{\epsilon_1, \epsilon_2\}$, we conclude that the rectangle $I \times J=[a-\epsilon, a+\epsilon] \times [b-\delta, b+\delta]$ has the desired property.
(2) I think I've managed to do the second part now. My reasoning was as follows: Let $I \times J$ be as above. For each $x \in I$, we have $f(x, b-\delta)<c<f(x, b+\delta)$, so, by the Intermediate Value Theorem, there exists some $\psi(x) \in (b-\delta, \delta)$ for which $f(x, \psi(x))=c$. We can then construct $\psi:I \rightarrow J$ by letting $\psi(a)=b$ and letting $\psi(x)$ be the real number in $J$ for which $f(x, \psi(x))=c$. We now have to prove that any such function is continuous at $x=a$. To do so, let $(x_n) \subset I$ be a sequence whose limit is $a$. Because $(\psi(x_n)) \subset J$, the only way for $(\psi(x_n))$ not to converge is if it had two converging subsequences with different limits. We prove that this is impossible by showing that whenever $(\psi(x_n))$ converges then its limit is $\psi(a)=b$. To this let, suppose that $\psi(x_n) \rightarrow d \in J$. The continuity of $f$ then gives us $$c=\lim c = \lim f(x_n, \psi(x_n))=f(\lim x_n, \lim \psi(x_n))=f(a, d).$$ However, we defined $J$ in such a way that $f(a, x)$ is increasing on it, so the only way for $f(a, d)=c$ is if $\lim \psi(x_n)=b=\psi(a)$. We thus proved that the function $\psi$ exists and is always continuous at $x=a$.
(3) I (think that) I proved (1) and (2). It now remains to show continuity and uniqueness for $\psi$ when $\frac{\partial f}{\partial y}>0$ everywhere. We first show continuity. Let $x \in I$ be arbitrary, and let $(x_n) \subset I$ be any sequence converging to $x$. Without loss of generality, we may assume $(\psi(x_n))$ converges, say, to $d$, so it remains to show that $d=\psi(x)$. Notice, however, that
$$f(x, \psi(x))=c = \lim c = \lim f(x_n, \psi(x_n)) = f(\lim x_n, \lim \psi(x_n))=f(x, d),$$ and, since $\partial f/\partial y>0$ everywhere, the only way for us to have $f(x, \psi(x))=f(x, d)$ is if $\psi(x)=d=\lim \psi(x_n)$, that is, $\psi$ is continuous at $x \in I$. Of course, the same argument applies to every $x \in I$, so we conclude $\psi$ is continuous.
Once we have showed continuity, we now show uniqueness. To do that, suppose that we have functions $\psi_1, \psi_2:I \rightarrow J$ such that $\psi_1(a)=\psi_2(a)=b$ and $f(x, \psi_1(x))=f(x, \psi_2(x))=c$ for every $x \in I$. Then let $x \in I$ be arbitrary, and consider any sequence $(x_n) \subset I$ converging to $x$. By continuity of $f$ and $\psi_i$, we have that $$c=\lim c=\lim f(x_n, \psi_1(x_n))=f(\lim x_n, \lim \psi_1(x_n))=f(x, \psi_1(\lim x_n))=f(x, \psi_1(x)).$$ Similarly, $c=f(x, \psi_2(X))$, from which it follows that $f(x, \psi_1(x))=f(x, \psi_2(x))$. However, for fixed $x$, the function $g$ defined by $g(y)=f(x, y)$ is increasing since $\partial f/\partial y>0$ everywhere, so the only way for us to have $f(x, \psi_1(x))=f(x, \psi_2(x))$ is if $\psi_1(x)=\psi_2(x)$. Since $x \in I$ was taken arbitrarily, uniqueness follows.