The function is
$$f(x) = \| x x^T - V \|_*$$
where $\| \cdot \|_*$ denotes the nuclear norm and $V$ is a given matrix. $x$ is a vector. Please tell me how to differentiate $f(x)$. And, if it is possible, please show me how to compute the 2nd derivative of $f(x)$.
Define a new matrix variable $$M=xx^T-V$$Then find the differential of the function in terms of this new variable $$\eqalign{ f &= \operatorname{tr}\sqrt{M^TM} \cr \cr df &= \frac{1}{2}(M^TM)^{-1/2}:d(M^TM) \cr &= \frac{1}{2}(M^TM)^{-1/2}:(dM^TM+M^TdM) \cr &= (M^TM)^{-1/2}:M^TdM \cr &= M(M^TM)^{-1/2}:dM \cr &= M(M^TM)^{-1/2}:d(xx^T) \cr &= M(M^TM)^{-1/2}:(dx\,x^T+x\,dx^T) \cr &= \big(M(M^TM)^{-1/2} + (M^TM)^{-1/2}M^T\big)\,x:dx \cr \cr \frac{\partial f}{\partial x} &= \big(M(M^TM)^{-1/2} + (M^TM)^{-1/2}M^T\big)\,x \cr\cr }$$ This is the gradient. To find the hessian you must differentiate again wrt $x$. But it's going to be very messy since each term $M$ contains two $x$'s inside of it.
And we can't use the "trace trick" again $$\operatorname{tr}(f(X))= f^\prime(X^T):dX$$ since there are no traces left in the gradient.
If there is other information that would simplify the problem (e.g. $V$ is symmetric), then you might be able to find an explicit formula for the hessian.
If you want the hessian in order to use something like Newton's method, I would suggest that you try a gradient-based method instead.
Since $V\!=\!V^T\,$ the Matrix Sign function can be used to write the gradient $$ \frac{\partial f}{\partial x} = 2\,\operatorname{sign}\left(xx^T-V\right)\,x $$
