The matrix $I_n - v^t x$ is invertible when $\langle v, x \rangle \neq 1$ 
Let $v \neq 0$ be any vector in $\Bbb R^n$ and let, $U_v = \{x \in \Bbb R^n : \langle v,x \rangle \neq 1\}$. Then:
(i) Show that the matrix $I_n - v^t x$ is invertible $\forall x \in U_v$.
(ii) Compute the derivative of the map $f:U_v \to GL_n(\Bbb k)$ ($\Bbb k = \Bbb R \text{ or } \Bbb C$) defined by $$f(x)={(I-v^tx)}^{-1} .$$

My attempt:
(i) I have tried to do it using determinant but couldn't do it.
(ii) Observed $f=f_2 \circ f_1$, where $$f_1 : U_v \to GL_n(\Bbb R)$$ $$x \mapsto (I_n - v^t x)$$   and $$ f_2 : GL_n(\Bbb R) \to GL_n(\Bbb R)$$ $$ A \mapsto A^{-1}$$
Then derived, ${Df_1}_{(x)} (h)= -v^t h$ and ${Df_2}_{(A)}(H)= -A^{-1}HA^{-1} $.
Applied Chain Rule to obtain, $$Df_x (h)= {(I_n - v^tx)}^{-1} (v^t h){(I_n - v^tx)}^{-1}$$
I think I have done part (ii) only for $\Bbb k = \Bbb R$.
How to prove part (i) and if there is any mistake in my attempt in part $(ii)$, then please point it out. Thanks in advance for help!
 A: The part $(1)$ of your question follows from the well-known matrix determinant lemma (the proof of which is not hard).
Using the lemma, we get
$$
\det (I_n - v^t x) = (1-x^t I_n^{-1} v ) \det I_n = 1 - x^t v = 1- <x,v> \neq 0,
$$
hence the matrix in question is invertible.
A: (i) One option is to try and construct explicitly an inverse for $I_n - v^t x$. This is a modification of the identity matrix by a rank-$1$ matrix, so it's natural to guess that an inverse would have the same property: We thus write the candidate inverse as $I_n + w^t y$ for some vectors $w, y$ (all vectors here are row vectors). By definition, we have
$$I_n = (I_n - v^t x)(I_n + w^t y) .$$
Multiplying both sides on the right by $y^T$ shows that $v \parallel w$, and instead multiplying both sides on the left by $w$ shows that $x \parallel y$, so the inverse must have the form $$(I_n - v^t x)^{-1} = I_n + \lambda v^t x$$ for some $\lambda$. Using again the definition of matrix inverse and solving gives $\lambda$. (Note that the expression is undefined when $v x^t = 1$, which hints at why that condition was necessary.)
Remark This gives a special case of the Sherman-Morrison Formula, itself a special case of the Woodbury Matrix Identity.
Alternatively, one can compute the determinant of $I_n - v^t x$ using the Matrix Determinant Lemma, which gives the determinant of a rank-$1$ update of a matrix. An easy proof for the case here (where the starting matrix is $I_n$) is given here.
(ii) Your solution looks good to me.
A: Here is a different approach where you can find the inverse explicitly. But first, let me write $X=x^t$ and $V=v^t$ so that whereas  your $v$ and $x$ are row vectors, my $X$ and $V$ are column vectors. (This is just a stylistic preference.)
Let $\alpha$ be a number you're trying to find so that $I_n+\alpha VX^t$ is the inverse of $I_n-VX^t$. This form can be suggestive from either intuition (note $(1-z)(1+\alpha z)=1+\alpha z-z-\alpha z^2$ and you have a situation here $z^2=\beta z$ for some $\beta$; see below) or the Matrix Determinant Formula linked to in Hayk's answer. So let's try it
$$
(I_n-VX^t)(I_n+\alpha VX^t)=I_n+\alpha VX^t-VX^t-(\alpha X^tV)VX^t
$$
so you want
$$
0=\alpha(1-X^tV)-1\implies\alpha=\frac{1}{1-X^tV}.
$$
Note we have used $X^tV\neq 1$. All of this guess work can now be made rigorous by verifying directly that
$$
(I_n-VX^t)\left(I_n+\frac{1}{1-X^tV}VX^t\right)=I_n=\left(I_n+\frac{1}{1-X^tV}VX^t\right)(I_n-VX^t).
$$
In your original notation, the last line above amounts to
$$
(I_n-v^tx)\left(I_n+\frac{1}{1-\langle x,v\rangle}v^tx\right)=I_n=\left(I_n+\frac{1}{1-\langle x,v\rangle}v^tx\right)(I_n-v^tx).
$$
