The Question: Suppose $f$ is differentiable at $x_\circ$ and $f^\prime(x_\circ) < 0$. Show that there exists a $\delta > 0$ such that $f(x) > f(x_\circ)$ for all $x \in (x_\circ - \delta, x_\circ)$ and that $f(x) < f(x_\circ)$ for all $x \in (x_\circ, x_\circ + \delta)$.
Hint: Note that it may be that $f$ is differentiable only at $x_\circ$ , or $f$ is differentiable on $\mathbb R$ but $f$ is not decreasing on any interval containing $x_\circ$. Use the definition of the derivative and the fact that if the limit $L$ of a function $g(x)$ at $x_\circ$ is nonzero, then $|g(x)| > \frac {|L|}{2}$ (which you proved earlier).
An attempt at an answer: By our assumption $f$ is differentiable at $x_\circ$, so for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $$ \left|\frac{f(x) - f(x_\circ)}{x-x_\circ} - f^\prime(x_\circ)\right| < \varepsilon \quad\text{when}\quad 0 < |x - x_\circ| < \delta\,. $$ Since we are also assuming that $f^\prime(x_\circ) < 0$, it is especially true that $f^\prime(x_\circ) \neq 0$, so by our previous result we know that \begin{equation} \left|\frac{f(x) - f(x_\circ)}{x-x_\circ}\right| > \frac{|f^\prime(x_\circ)|}{2}\,. \end{equation} Now by the triangle inequality $|x - y| \leq |x| + |y|$ and our previous result, \begin{aligned} \left|\frac{f(x) - f(x_\circ)}{x-x_\circ} - f^\prime(x_\circ) \right| &\leq \left|\frac{f(x) - f(x_\circ)}{x-x_\circ} \right| +\left| f^\prime(x_\circ)\right|\\ &= \left|\frac{f(x) - f(x_\circ)}{x-x_\circ} \right| +2\frac{\left|f^\prime(x_\circ)\right|}{2}\\ &< 3\left|\frac{f(x) - f(x_\circ)}{x-x_\circ} \right|\,. \end{aligned} Now this is where I'm stuck. Usually at this point I would have run into a situation, where evaluating things upwards would have resulted in an expression with an absolute value $|x - x_\circ|$ times some other expression, which I could have evaluated in one direction or the other. Obviously that is not the case here. How could one proceed in a situation like this, where there is no obvious expression $< \delta$?
Hint: Get rid of the absolute value.
For $|x - x_0| < \delta$ we have
$$f'(x_0) - \epsilon < \frac{f(x) - f(x_0)}{x - x_0} < f'(x_o) + \epsilon.$$
Take $\epsilon = - \frac{1}{2}f'(x_0)$ and work with the right-side inequality.