Step 1: Examine the inside of the greatest integer function.
As $x \to -\frac{3}{5}$, compute $\dfrac{-1}{x}$. Substituting the limiting value, $\dfrac{-1}{-\frac{3}{5}} = \dfrac{5}{3} \approx 1.667$.
Step 2: Check the behaviour near that value.
For $x$ close to $-\frac{3}{5}$, the quantity $\dfrac{-1}{x}$ stays a little above or below $\frac{5}{3}$, but it remains strictly between $1$ and $2$.
Step 3: Evaluate the greatest integer part.
Since $\dfrac{-1}{x}$ lies in $(1,2)$ throughout a small neighbourhood, $\left[\dfrac{-1}{x}\right] = 1$ there.
Step 4: Replace the bracket by its constant value.
The expression becomes $\dfrac{1}{x}\cdot 1 = \dfrac{1}{x}$ near the limit point.
Step 5: Take the limit of $\dfrac{1}{x}$.
As $x \to -\frac{3}{5}$, $\dfrac{1}{x} \to \dfrac{1}{-\frac{3}{5}} = -\dfrac{5}{3}$.
Step 6: State the result.
Hence the limit equals $-\dfrac{5}{3}$.
\[ \boxed{-\dfrac{5}{3}} \]