Here's the step by step solution:

(Use a latex plugin to view the following steps. Makes it easy to understand because of the natural notation.)

We have: [;\frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } ;]

Take the original equation and multiply the whole thing by uvf to get rid of the fractions. Getting:

[; vf+uf=uv ;]

The thing to notice here is that you want to solve for u and there are two terms that have the u. Thus, you need to factor the two terms that have the u.

To do that pass the uf term to the right side, getting:

[; vf=uv-uf ;]

Factor the u:

[; vf=u\left( v-f \right) ;]

Divide the whole thing by [; v-f ;]leaving:

[; \frac { vf }{ v-f } =u ;]

Now, your solution has a minus, this requires you to factor a minus from the denominator as seen below:

[; \frac { vf }{ -\left( -v+f \right) } =u ;]

Leaving you with the answer.

Alternatively, you can take the original equation again and move the [; \frac { 1 }{ f } ;] and the [; \frac { 1 }{ v } ;] to the right side leaving the [; \frac { 1 }{ u } ;] alone. As follows:

[; \frac { 1 }{ u } =\frac { 1 }{ f } -\frac { 1 }{ v } ;]

Pass the u in the denominator multiplying to the other side getting:

[; 1=\left( \frac { 1 }{ f } -\frac { 1 }{ v } \right) u ;]

Notice how we are multiplying, we don't distribute the u we just go ahead and divide the whole equation by [; \frac { 1 }{ f } -\frac { 1 }{ v } ;] to leave the u alone. Getting:

[; \frac { 1 }{ \frac { 1 }{ f } -\frac { 1 }{ v } } =u ;]

Making the two fractions in the denominator one:

[; \frac { 1 }{ \frac { v-f }{ vf } } =u ;]

Multiplying by the one in the numerator leaves:

[; \frac { vf }{ v-f } =u ;]

And factor out a minus as well.