Answer
$$L(x) =-0.002 x+0.03$$$0.0027\%$
Work Step by Step
Given $$\frac{1}{(10.03)^{2}} $$
Consider $f(x)=\dfrac{1}{x^2}, a=10,$ and $\Delta x=0.03$, since
\begin{align*}
f^{\prime}(x)&=-2 x^{-3}\\
f^{\prime}(10)&=-0.002
\end{align*}
Then the linearization to $f (x)$ is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&= -0.002(x-10)+0.01\\
&=-0.002 x+0.03
\end{align*}
Since
\begin{align*}
L(10.03)&= f(10.03)\\
&=-0.002(10.03)+0.03\\
&\approx 0.00994
\end{align*}
Hence the error is given by
$$ | \frac{1}{(10.03)^2}-0.00994|= 2.69\times 10^{-7}$$
and the percentage is
$$\frac{2.97\times 10^{-5}}{\frac{1}{(10.03)^2}}\times 100 \% \approx 0.0027\%$$
