$f(\mathbf{x}) = \frac{1}{p}\sum_{i=1}^n |x_i|^p$, for $1 < p < \infty$. Find the conjugate of $f$, $f^*$.
My attempt: \begin{align} f^*(y) &= sup_{x \in \mathbb{R}^n} \{y^\intercal x - \frac{1}{p}\sum_{i=1}^n |x_i|^p\}\\ &= sup_{x \in \mathbb{R}^n} \{\|x\|_p\|y\|_q - \frac{1}{p}\|x\|_p^p\} &\text{Holder's Inequality: } y^\intercal x \leq \|x\|_p\|y\|_q \text{, where } \frac{1}{p} + \frac{1}{q} = 1\\ &= sup_{x \in \mathbb{R}^n} \{\|x\|_p(\|y\|_q - \frac{1}{p}\|x\|_p^{p-1})\}\\ &= \begin{cases} 0, & \text{if $\|y\|_q \leq \frac{1}{p}\|x\|_p^{p-1}$}.\\ \infty, & \text{otherwise}. \end{cases} \end{align}
I am not too sure if I am in the right direction. It looks incorrect because of the last line where $y$ is still depending on $x$?
Your last line is not correct. You have to maximize $a(b-\frac 1 p a^{p-1})$ over all $a \geq 0$ where $b=\|y\|_q$. The derivative w.r.t. $a$ is $(b-\frac 1 p a^{p-1})-a\frac 1 p (p-1)a^{p-2}=b-a^{p-1}$. The maximum value is attained when $a=b^{\frac 1 {p-1}}$. Luckily $x_i=sgn(y_i) |y_i|^{q/p}$ gives a vector where this property holds and you also have equality in Holder's inequality in your earlier step. So this choice of $x$ indeed gives you the value of $f^{*}(y)$.
[Here $sgn (y_i) =1$ if $y_i \geq 0$ and $-1$ if $y_i <0$].