Firstly we should notice the similarities between this expression and expressions of the standard type ax.
To deal with this type of expression, we take logarithms of both sides and then apply the usual rules of differentiation. Let's try that here:
1.) y = xx
take logs of both sides and use the multiplicative law of logarithms
2.) log(y) = xlog(x)
now use the chain rule on the LHS and the product rule on the RHS
3.) dy/dx * 1/y = 1 + log(x)
now simply rearrange to find dy/dx
4.) dy/dx = xlog(x)*(1+log(x))