Solutions

a) We have

$$\begin{align*} f'(a)&=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} \\ &=\lim_{h\to 0}\frac{\frac{1}{(a+h)^2}-\frac{1}{a^2}}{h} \\ &=\lim_{h\to 0}\frac{a^2-a^2-2ah-h^2}{h(a+h)^2a^2} \\ &=\lim_{h\to 0}\frac{-h-2a}{(a+h)^2a^2} \\ &=\frac{-2a}{a^4} \\ &=-\frac{2}{a^3}. \end{align*}$$

b) The tangent line is defined as

$$y=-\frac{2}{a^3}(x-a)+\frac{1}{a^2}\implies y=-\frac{2x}{a^3}+\frac{3}{a^2}.$$

The intersection is at

$$\begin{align*} -\frac{2x}{a^3}+\frac{3}{a^2}&=\frac{1}{x^2} \\ &\implies -\frac{2x^3}{a^3}+\frac{3x^2}{a^2}=1 \\ &\implies \frac{2x^3}{a^3}-\frac{3x^2}{a^2}+1=0 \\ &\implies 2x^3-3ax^2+a^3=0 \\ &\implies (x-a)(2x^2-xa-a^2)=0 \\ &\implies (x-a)^2(2x+a)=0 \\ &\implies x=a\quad\text{or}\quad x=-\frac{a}{2}. \end{align*}$$

Since one of the solutions is $x=-\frac{a}{2}$, $f$ intersects the tangent line on the opposite side of the vertical axis.