Absolute error in approximation of product Q: Suppose $x_t,y_t$ are the true values with approximations $x_a,y_a$ respectively with small absolute errors $e_x,e_y$. Show that the absolute error in the approximation of their product $e_{xy}$ satisfies the following upper bound:

$$e_{xy}\leq |x_ay_a|\left(\left|\frac{e_x}{x_a}\right|+\left|\frac{e_y}{y_a}\right|\right)$$

One way I know is relying on a theorem saying that for a function $f(x,y)$ partially differentiable w.r.t its arguments, we have $$e_{f}\leq\left|\frac{\partial f}{\partial x}e_x\right|+\left|\frac{\partial f}{\partial y}e_y\right|$$
at $(x,y)=(x_a,y_a)$
which directly gives that $e_{xy}\leq |y_ae_x|+|x_ae_y|=|x_ay_a|\left(\left|\dfrac{e_x}{x_a}\right|+\left|\dfrac{e_y}{y_a}\right|\right)$ since $f(x,y)=xy$ is partially differentiale w.r.t its arguments.

What I want to do is to prove this without using that theorem and just using the basic definitions. How should I go about it? I've tried manipulating it using the triangle inequality (also the reverse analogue of it) but I can't quite get there. 
Can someone give me a hint for a viable way to proceed? Please refrain from posting full solutions though. I'm preparing for an exam, so I need to work it out myself.
I'm using $e_x=|x_t-x_a|$, $e_y=|y_t-y_a|$ and $e_{xy}=|x_ty_t-x_ay_a|$.
Thanks!
 A: The problem is wrong. These things happen. A simple counter example has been supplied by Lutz, i.e. $x_t = y_t = 1$ and $x_a = y_a = 1 + \epsilon$ for any $\epsilon > 0$. Then $e_{xy} = 2 \epsilon + \epsilon^2$, while the proposed upper bound is $2 \epsilon$. 
Since your problems description is wrong, I will supply you with a complete analysis rather than the requested guidance.
Our target is $T = x_t y_t$. The approximation is $A = x_a y_a$. The error $E = T - A$ is
\begin{align} 
E &= x_t y_t - x_a y_a \\ &= x_t y_t - x_t y_a + x_a y_a - x_a y_a \\&= x_t (y_t - y_a) - (x_t - x_a) y_a \\&= x_t e_y - e_x y_a
\end{align}
To eliminate $y_a$ write
$$ y_a = y_t - (y_t - y_a) = y_t - \frac{y_t - y_a}{y_t} y_t = y_t (1 - r_y)$$
where $r_y$ is the relative error $$r_y = \frac{y_t - y_a}{y_t} $$
You can then estimate
\begin{align}
|E| 
&\leq |x_ty_t| \left( \frac{|e_y|}{|y_t|} + \frac{|e_x|}{|x_t|}(1 + |r_y|)\right) \\&= |x_ty_t| ( |r_y| + |r_x|(1+|r_y|) \\
&= |x_ty_t| (|r_x| + |r_y| + |r_x||r_y|)
\end{align}
The example quoted above where $|r_x| = |r_y| = \epsilon$ shows that equality is possible.
