a. Recall that for [tex]|x|<1[/tex], we have
[tex]\dfrac1{1-x}=\displaystyle\sum_{n\ge0}x^n[/tex]
which has derivative
[tex]\dfrac1{(1-x)^2}=\displaystyle\sum_{n\ge0}nx^{n-1}=\sum_{n\ge0}(n+1)x^n[/tex]
Then
[tex]f(x)=\dfrac1{(10+x)^2}=\dfrac1{100}\dfrac1{\left(1-\left(-\frac x{10}\right)\right)^2}=\frac1{100}\displaystyle\sum_{n\ge0}(n+1)\left(-\frac x{10}\right)^n[/tex]
which converges
[tex]\left|-\dfrac x{10}\right|<1\implies|x|<10[/tex]
b. From the above result, it's evident that the radius of convergence is [tex]R=10[/tex].