Backward stability theorem 
Theorem. Let $\tilde{f}(x)$ be a backward stable algorithm of $f(x)$, with a relative condition number $\kappa(x)$. Then $${{\Vert \tilde{f}(x) - f(x) \Vert}\over{\Vert f(x) \Vert}} \leq \kappa (x) \varepsilon_m$$ Where $\varepsilon_m$ is a machine number

So, a backward stable algorithm $\tilde f$ is one such that $\tilde f(x) = f(\tilde x)$ for each $x$ such that ${\Vert \tilde x - x \Vert \over \Vert x \Vert} = \mathcal{O}(\varepsilon_m)$.
I'm thinking If $\tilde f$ is backward stable, then with $\tilde x = x(1+\epsilon_1)$ we have $$\tilde f(x) = f(x(1+\epsilon_1))$$
Given that for each $x$, ${\Vert \tilde x - x \Vert \over \Vert x \Vert} = \mathcal{O}(\varepsilon_m)$, we have \begin{align*} {\Vert \tilde x - x \Vert \over \Vert x \Vert } 
 &= {\Vert x(1+\epsilon_1)-x \Vert \over \Vert x \Vert} \\ &= {\Vert x \epsilon_1 \Vert \over \Vert x \Vert} \\ &\leq {\Vert \epsilon_1 \Vert} \end{align*}
And \begin{align*} {\Vert \tilde f(x) - f(x) \Vert \over \Vert f(x) \Vert} &= {\Vert f(\tilde x) - f(x) \Vert \over \Vert f(x) \Vert}\end{align*} but at this point I don't know how to continue. Where do I go from here? The definition I have for the condition number is
$$ \kappa (x) = \lim_{\epsilon \rightarrow 0}\biggr( \sup_{|\delta x| \leq \epsilon}\biggr| {\delta f / f \over\delta x / x} \biggr|\biggr)$$
 A: As the theorem is currently stated it is not entirely true and I feel that the notation is a bit misleading. In particular, the distinction between normwise and componentwise errors is not emphasized.
I shall derive the correct theorem in the case of functions that are normwise relative backward stable. To that end we must first define the corresponding condition number.
We consider a function $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ and a point $x \in \mathbb{R}^n$ such that $x \not = 0$ and $f(x) \not = 0$. Our objective is to quantify how sensitive the function $f$ is to small changes in the input argument $x$. We begin by defining the auxiliary function $\kappa_f(x,\epsilon)$ given by
$$ \kappa_f(x,\epsilon) = \sup \left\{ \left(\frac{\|f(x) - f(y)\|}{\|f(x)\|}\right) \big/ \left( \frac{\|x-y\|}{\|x\|} \right) \: : \: \|x-y\| \leq \epsilon \|x\|\right\}. $$
We see here why it is necessary to assume that $x \not =0$ and $f(x) \not =0$.
It clear that $\epsilon \rightarrow \kappa_f(x,\epsilon)$ is an increasing and nonnegative function of $f$. It follows that the limit
$$ \underset{\epsilon \rightarrow 0_+}{\lim} \kappa_f(x,\epsilon)$$ actually exists. We define the normwise condition number of the function $f$ at the point $x$ as this limit, i.e.,
$$ \kappa_f(x) = \underset{\epsilon \rightarrow 0_+}{\lim} \kappa_f(x,\epsilon).$$
The normwise condition number imposes a hard limit on the accuracy that can be achieved. In particular, if  $y \in \mathbb{R}^n$ is any point such that $\|x-y\| \leq \epsilon \|x\|$, then by definition, we have
$$ \frac{\|f(x) - f(y)\|}{\|f(x)\|} \leq \kappa_f(x, \epsilon) \frac{\|x-y\|}{\|x\|}.$$
It follows that we cannot expect to have a relative error that is smaller than
$$ \frac{\|f(x) - f(y)\|}{\|f(x)\|} \approx \kappa_f(x) \frac{\|x-y\|}{\|x\|}.$$
Now let $u$ denote the unit roundoff. Let us consider an algorithm for computing $$z = f(x)$$ that is normwise relative backward stable. Specifically, it has the following property
$$\forall x \in \mathbb{R}^n  \setminus \{0\} \: \exists C > 0 \: \exists\,y \in \mathbb{R}^n \: : \: \hat{z} = f(y) \: \wedge \: \frac{\|x-y\|}{\|x\|} \leq C u.$$
Here $\hat{z}$ denotes the computed value of $z = f(x)$, i.e., the value that is returned by the computer. Small values of $C$ correspond to well-designed algorithms that are implemented by skillful programmers who are familiar with finite precision arithmetic. What can be said about the relative forward error given by $$\frac{\|z-\hat{z}\|}{\|z\|}.$$ How can it be bounded from above?
We have
$$\frac{\|z-\hat{z}\|}{\|z\|} = \frac{\|f(x)-f(y)\|}{\|f(x)\|} \leq \kappa_f(x,\delta) \frac{\|x-y\|}{\|x\|}, $$
where $\delta > 0$ is any number such that
$$\|x-y\| \leq \delta \|x\|$$
In particular, we are free to choose $\delta = Cu$. It follows that
$$ \frac{\|f(x)-f(y)\|}{\|f(x)\|} \leq \kappa_f(x,Cu) Cu $$
Now if $Cu$ is sufficiently small, then
$$ \kappa_f(x) \approx \kappa_f(x,Cu)$$
is a good approximation and we may write
$$ \frac{\|f(x)-f(y)\|}{\|f(x)\|} \lesssim \kappa_f(x) Cu, $$
but that is as far as we can go. In particular, we cannot say with certainty that
$$ \frac{\|f(x)-f(y)\|}{\|f(x)\|} \leq \kappa_f(x) Cu. $$
The original problem description suggested that $C = 1$. That is a very good value that is hard to achieve in practice.
