I know that the Fenchel conjugate of a function is $$f^*(x^*) = \sup_x\{\langle x, x^*\rangle - f(x)\}.$$ However, how do I find the Fenchel conjugate of the function $$f(x) = \frac{1}{p}\sum\limits_{i=1}^n |x_i|^p$$ where $1 < p < \infty$.
I have tried differentiating the equation and taking it to be $= 0$ but I cannot seem to reach any answer.
It will be, somewhat expectedly, $$f^*(x^*) = \frac{\|x^*\|_q^q}{q}, \quad \frac1q + \frac1p = 1$$ as indicated in the comments to the OP. The paper linked there was already rightfully identified as sketchy, the derivation there is wrong on several levels, which is why I am including this answer.
I will use $q$ for the conjugated exponent to $p$ in the following without mentioning this explicitly. Let $$G(x) := \langle x^*,x \rangle - f(x).$$ First note that $\frac1q \|x^*\|_q^q$ is an upper bound on $G$ since $$\langle x^*,x\rangle \leq \frac{\|x^*\|_q^q}{q} + \frac{\|x\|_p^p}{p}.$$ Thus, if a stationary point $\bar x$ of $G$ realizes this upper bound, we are done.
Now, $$\partial_j G(x) = x^*_j - |x_j|^{p-2}x_j,$$ hence $\nabla G(x) = 0$ if and only if $x^*_j = |x_j|^{p-2}x_j$ for each $j$. (The linked paper gets this wrong and then proceeds to mix up all norms.) It is easy to see that $\alpha(t) = |t|^{p-2}t$ and $\omega(s) := |s|^{q-2}s$ are bijective functions on $\mathbb{R}$ which are inverse to each other. Hence $x^*_j = |x_j|^{p-2}x_j$ if and only if $x_j = |x^*_j|^{q-2}x_j^*$. So this is the designated maximizer which we call $\bar x$. There holds $\|\bar x\|_p^{q-1} = \|x^*\|_q$ and thus indeed $$G(\bar x) = \|x^*\|_q^q - \frac1p \|x^*\|_q^q = \frac{\|x^*\|_q^q}q.$$
