Once you have an equation involving $\frac{\nu}{s}$ and $\frac{\nu}{t}$, what can you multiply the whole equation by so that $s$ and $t$ no longer appear in denominators?

After clearing denominators, you will get an equation where $s$ appears in more than one term. Put all the terms with $s$ on one side, factor out $s$, and then divide to solve for $s$.

Because algebra is the faster primary method, use Desmos to verify the choices. Enter t=3 and v=2, using v to represent $\nu$.

Use valid single-letter variables for the four choices. Enter a=(2t-v)/(v*t), b=v/(2t-v), c=2v/(t-v), and d=(v*t)/(2t-v).

On separate lines, enter 2/(1/a+1/t), 2/(1/b+1/t), 2/(1/c+1/t), and 2/(1/d+1/t). Compare each result with v; only the expression that evaluates to 2 satisfies the original equation for these values.

Begin with

Multiply both sides by the denominator $\frac{1}{s} + \frac{1}{t}$ to get rid of the fraction in the denominator:

From

distribute $\nu$:

Now multiply the entire equation by $st$ (the common denominator of $s$ and $t$) to clear the remaining fractions:

We now have

Move the $\nu s$ term to the right side so that all $s$ terms are together:

Factor $s$ out of the terms on the right:

From

divide both sides by $2t - \nu$ (which is allowed as long as $2t - \nu \neq 0$):

So $s$ expressed in terms of $t$ and $\nu$ is $s = \dfrac{\nu t}{2t - \nu}$, which corresponds to choice D.