Let's use a colon to denote the trace/Frobenius product, i.e.
$$A:B={\rm tr}(A^TB)$$
and define a new (symmetric) matrix variable
$$M=xx^T$$
We'll also need a general result about the differential of the trace of a function
$$d\,{\rm tr}(f(X))=f'(X^T):dX$$
where $f'$ denotes the derivative of $f$.
Write the function in terms of this new variable and find its differential and then its gradient
$$\eqalign{
f &= {\rm tr}\big(M\exp(M)\big) \cr\cr
df &= (I+M)\exp(M):dM \cr
&= (I+M)\exp(M):(dx\,x^T+x\,dx^T) \cr
&= 2\,(I+M)\exp(M):dx\,x^T \cr
&= 2\,(I+M)\exp(M)x:dx \cr\cr
\frac{\partial f}{\partial x} &= 2\,(I+xx^T)\exp(xx^T)x \cr\cr
}$$
Let's use a colon to denote the trace/Frobenius product, i.e. $$A:B={\rm tr}(A^TB)$$ and define a new (symmetric) matrix variable $$M=xx^T$$ We'll also need a general result about the differential of the trace of a function $$d\,{\rm tr}(f(X))=f'(X^T):dX$$ where $f'$ denotes the derivative of $f$.
Write the function in terms of this new variable and find its differential and then its gradient $$\eqalign{ f &= {\rm tr}\big(M\exp(M)\big) \cr\cr df &= (I+M)\exp(M):dM \cr &= (I+M)\exp(M):(dx\,x^T+x\,dx^T) \cr &= 2\,(I+M)\exp(M):dx\,x^T \cr &= 2\,(I+M)\exp(M)x:dx \cr\cr \frac{\partial f}{\partial x} &= 2\,(I+xx^T)\exp(xx^T)x \cr\cr }$$