$f$ is continuous $ \iff $ $f^{-1}$ is continuous?

It is not true in general. For example, consider $A = [0, 1) \cup [2, 3]$ and $f(x) = x$ on $[0, 1)$ and $f(x) = x - 1$ on $[2, 3]$. Clearly, $f$ is continuous and bijective onto $[0, 2]$. The inverse function $g\colon [0, 2] \to A$ is defined by $g(x) = x$ for $x \in [0, 1)$ and $g(x) = x + 1$ for $x \in [1, 2]$. Clearly, $g$ is not continuous.

If you require that $A$ and $B$ are intervals, then the assertion is correct. To prove this, let $f\colon A \to B$ be continuous and bijective (where $A$ and $B$ are intervals). Then $f$ is strictly monotonous, which can easily be shown using the intermediate value theorem. But the inverse of a strictly monotonous function on an interval is continuous, which proves the assertion.

I can elaborate more on that, if you want.

In general it is not true, but you can show that (supposing $f$ continuos) for every point in which $\lim_{x\to x_0} f^{-1} (x)$ exists, then it equals $f^{-1}(x_0)$.

In fact,

Let $f$ be continuos.

We want to prove $\lim_{x \to x_0} f^{-1}(x) = f^{-1}(x_0)$.

Now, $\lim_{x \to x_0} f^{-1}(x) = (f^{-1} \circ f)(\lim_{x \to x_0} f^{-1}(x))$

But given that $f$ is continuos, we have $f^{-1}( \lim_{x \to x_0} f(f^{-1}(x)) = f^{-1}( \lim_{x \to x_0} x) = f^{-1}(x_0) $ and we are done.