Don't Start Hating The L'Hopital's Rule While Solving This Problem! (Part 1)

$= \large \displaystyle \lim_{x \to 0} \frac{a^x - a^{\frac{x}{x+1}}}{x^2}$

$= \large \displaystyle \lim_{x \to 0} \frac{a^{\frac{x}{x+1}}\left(a^{x-\frac{x}{x+1}}-1\right)}{x^2}$

$= \large \displaystyle a^{\frac{0}{0+1}} \times \lim_{x \to 0} \frac{a^{\frac{x^2}{x+1}}-1}{x^2}$

$= \large \displaystyle 1 \times \lim_{x \to 0} \frac{\left(a^{\frac{x^2}{x+1}}-1\right)}{\left(\frac{x^2}{x+1}\right)}\cdot \frac{1}{x+1}$

$= \large \displaystyle \lim_{x \to 0} \frac{\left(a^{\frac{x^2}{x+1}}-1\right)}{\left(\frac{x^2}{x+1}\right)} \cdot \lim_{x \to 0} \frac{1}{x+1}$

$= \large \displaystyle \ln a \cdot \frac{1}{0+1} = \boxed{\ln a}$

Relevant wiki: Repeated Application of L'Hopital's Rule - Medium

$\begin{aligned} L & = \lim_{x \to 0} \frac {a^x-a^\frac x{x+1}}{x^2} & \small {\color{#3D99F6}\text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \lim_{x \to 0} \frac {\ln a \ a^x - \ln a \ a^\frac x{x+1} \cdot \frac 1{(x+1)^2}}{2x} & \small {\color{#3D99F6}\text{Differentiate up and down w.r.t. }x} \\ & = \lim_{x \to 0} \frac {\ln^2 a \ a^x - \ln^2 a \ a^\frac x{x+1} \cdot \frac 1{(x+1)^4} - \ln a \ a^\frac x{x+1} \cdot \frac {-2}{(x+1)^3}}{2} & \small {\color{#3D99F6}\text{Differentiate again}} \\ & = \frac {\ln^2 a - \ln^2 a + 2\ln a}{2} \\ & = \boxed{\ln a} \end{aligned}$

Use Maclaurin's series of a^x that will reduce work in differentiating twice