Step 1: Recognise the inner limit as a derivative.
Look at $\lim_{y\to 2}\frac{1}{y-2}\left(\frac{1}{x}-\frac{1}{x+y-2}\right)$. If we set $g(y)=-\frac{1}{x+y-2}$, the bracket is $g(y)-g(2)$ divided by $y-2$, which is exactly the definition of $g'(2)$.
Step 2: Differentiate $g$ with respect to $y$.
Treating $x$ as constant, $g(y)=-(x+y-2)^{-1}$, so $g'(y)=(x+y-2)^{-2}$.
Step 3: Evaluate the inner limit.
Putting $y=2$ gives $g'(2)=\frac{1}{(x+2-2)^2}=\frac{1}{x^2}$. So the quantity inside the outer derivative simplifies to $\frac{1}{x^2}$.
Step 4: Set up the outer derivative.
We must now compute $\frac{d}{dx}\left(\frac{1}{x^2}\right)$, where $x$ is the genuine variable of differentiation.
Step 5: Write as a power and differentiate.
Since $\frac{1}{x^2}=x^{-2}$, the power rule gives $\frac{d}{dx}\left(x^{-2}\right)=-2x^{-3}$.
Step 6: State the final value.
Therefore the whole expression equals $-\frac{2}{x^3}$, agreeing with the key. \[ \boxed{-\dfrac{2}{x^3}} \]