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.



Finding the conjugate of a function
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
$$G(\bar x) = \|x^\*\|\_q^q - \frac1p \|x^\*\|\_q^q = \frac{\|x^\*\|\_q^q}q.$$ 


