You want to find the difference quotient given that $\displaystyle f(x)=\dfrac{1}{x}$. It follows that $\displaystyle f(x+h)=\dfrac{1}{x+h}$.
Therefore,
$\displaystyle \begin{aligned}\dfrac{f(x+h)-f(x)}{h}&=\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}\\&=\dfrac{x-(x+h)}{hx(x+h)}\\&=\ldots\end{aligned}$.
I leave it for you to finish the problem. Does this make sense though?
That was the result of multiplying by the LCD of the two denominators in the numerator of the complex fraction. The LCD would be x(x + h), so
$\displaystyle \begin{aligned}
\dfrac{f(x+h)-f(x)}{h}&=\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}\\
&= \dfrac{\frac{1}{x+h}-\frac{1}{x}}{h} \cdot \dfrac{x(x + h)}{x(x +h)} \\
&= \dfrac{\frac{x(x + h)}{x+h}-\frac{x(x + h)}{x}}{hx(x + h)} \\
&=\dfrac{x-(x+h)}{hx(x+h)}\\
&=\ldots\end{aligned}$