$A$ is some fixed matrix. Let $U(B)=AB-BA$. If $A$ is diagonalizable then so is $U$?

(a) True. Indeed $p(T)B = p(A)B$ for every polynomial in $F[X]$ whence $p(T) = 0$ if and only if $p(A)=0$. By the minimal polynomial characterization of diagonalizability in finite dimension (splits with simple roots), it follows that $T$ is diagonalizable if and only $A$ is diagonalizable.

(b) True. If $A=P^{-1}DP$ is diagonalizable, then $U$ is similar to $V(B)=DB-BD$ with $D=\mbox{diag}(d_1,\ldots,d_n)$ diagonal. Precisely, $U=S\circ V\circ S^{-1}$ with $S(B)=P^{-1}BP$.

Then denoting $E_{ij}$ the canonical basis of $M_n(F)$, we find $$V(E_{ij})=DE_{ij}-E_{ij}D=(d_i-d_j)E_{ij}.$$ Whence $V$ is diagonal in the canonical basis, and $U$ is therefore diagonalizable.

Diagonalise $A$ as $PDP^{-1}$. Then the matrix of $U$ w.r.t. the canonical basis of $M_n(F)$ is $I\otimes A - A^T\otimes I$. Therefore $$ (P^T \otimes P^{-1}) (I\otimes A - A^T\otimes I) (P^{-T} \otimes P) = I\otimes D - D\otimes I $$ is diagonal, i.e. $U$ is diagonalisable.

This answer, of course, requires knowledge in Kronecker product. As your question comes from a Hoffman and Kunze exercise, you are not expected to use Kronecker product, but to solve the question in a linear algebraic way. However, I think this exercise is a perfect example that sometimes a matrix theoretic solution can be much neater than a linear algebraic one.