Given $f(2)=1$, $f’(2)=3$, and $f’’(2)=e$, find the first and second derivative of $f^{-1}(x)$ at $x=1$. I was studying calculus and bumped into this exercise:

Let $f(x)$ be a differentiable, one-to-one function.
Suppose $f(2)=1$, $f’(2)=3$, and $f’’(2)=e$, find the first and second derivative of $f^{-1}(x)$ at $x=1$.

Here is my thought:
I think I should find out what $f(x)$ is, and compute it’s inverse function subsequently. However I couldn’t figure out what $f(x)$ could be. Also, I noticed that $f^{-1}(1)=2$, which may help me find the answer. Any suggestions?  Or am I overlooking something important? Thanks a lot.
 A: Hint:
Let $g(x)$ be the inverse function of $f(x)$. Note that $$g(f(x))=x$$ and apply Chain Rule twice to find $g'(1)$ and $g''(1)$ respectively.
Note: It's sometimes pretty hard to find out $f(x)$ when you are given $f(a)$, $f'(a)$ and $f''(a)$ instead of $f(a)$, $f(b)$ and $f(c)$. Therefore, it should be better to turn to the  original meaning of an inverse function.
A: hint
Let $$g(x)=f^{-1}(x)$$
we know that
$$g'(x)=\frac{1}{f'(g(x))}$$
thus
$$g'(1)=\frac{1}{f'(g(1))}$$
but
$$g(1)=2$$
and
$$f'(2)=3$$
then
$$g'(1)=\frac{1}{f'(2)}=\frac 13$$
Differentiating the formula above, we get
$$g''(x)=-f''(g(x))g'(x).(g'(x))^2$$
So,
$$g''(1)=-f''(g(1))(g'(1))^3$$
$$=-\frac{1}{27}f''(2)=-\frac{e}{27}$$
A: Here is an idea,  we know that by definition of inverse,
$$ f(f^{-1} (x) ) = x $$
Suppose we differentiated both sides using the chain rule on left,
$$ \frac{df(x)}{dx}|_{f^{-1} (x) } \frac{df^{-1} (x) }{dx} = 1 \tag{1}$$
Now, what would happen if you plugged $x=1$ into the equation above?
For the second derivative, let's differentiate our previously derived identity(1) once again with the product rule,
$$ \frac{d^2 f(x) }{dx^2}|_{f^{-1} (x) } ( \frac{df^{-1} (x) }{dx})^2 + \frac{df}{dx}|_{f^{-1}(x) } \frac{d^2 f^{-1} (x) }{dx^2} = 0 \tag{2}$$
And, we have related the inverse function's derivative with the function's derivatives
Hope this helps!

Let's do better! If we rearrange (1),
$$ \frac{df^{-1} (x) }{dx}  = \frac{1}{ \frac{df}{dx}|_{f^{-1} (x) } } $$
And we can plug the above identity into the second factor of first term in $(2)$ which leads us to:
$$ \frac{d^2 f(x) }{dx^2}|_{f^{-1} (x) } ( \frac{1}{ \frac{df}{dx}|_{f^{-1} (x) } } )^2 + \frac{df}{dx}|_{f^{-1}(x) } \frac{d^2 f^{-1} (x) }{dx^2} = 0 \tag{3}$$
Now we have expressed the inverse function's derivative completely in the derivatives of the function.
A: Let $y=f^{-1}(x)$. Then $f(y)=x$ so differentiating with respect to $x$, we find that $\frac{d}{dx}=\frac{d}{dy}\frac{dy}{dx}$. Hence, $f'(y)y'=1$ so $y'=\frac{1}{f'(y)}=\frac{1}{f'(f^{-1}(x))}$. From this result, it should be obvious how to get $(f^{-1}(x))'$ at $x=1$.
Differentiate again.
\begin{align}
y''&=-(f'(f^{-1}(x)))^{-2}[f'(f^{-1}(x))]'\\
&=-(f'(f^{-1}(x)))^{-2}f''(f^{-1}(x))y'.
\end{align}
Hence, you know everything. Just plug in.
