jmc

Since the remainder is $2x+1$ when $f(x)$ is divided by $(x-1{)}^2,$ we can write $$\begin{array}{rl}f(x)& =q(x)(x-1{)}^2+2x+1\\ & =q(x)(x-1{)}^2+2(x-1)+3.\end{array}$$Then $$\frac{f(x)-3}{x-1}=q(x)(x-1)+2.$$Let $$g(x)=q(x)(x-1)+2.$$By the Remainder Theorem, $f(3)=15,$ so $$g(3)=\frac{f(3)-3}{3-1}=\frac{15-3}{3-1}=6.$$Also, $g(1)=2.$

Let $ax+b$ be the remainder when $g(x)$ is divided by $(x-1)(x-3),$ so $$g(x)=q_1(x)(x-1)(x-3)+ax+b.$$Setting $x=1$ and $x=3$ gives $$\begin{array}{rl}a+b& =g(1)=2,\\ 3a+b& =g(3)=6.\end{array}$$Solving this system, we get $a=2$ and $b=0,$ so $$g(x)=q_1(x)(x-1)(x-3)+2x.$$Then $$\begin{array}{rl}f(x)& =g(x)(x-1)+3\\ & =[q_1(x)(x-1)(x-3)+2x](x-1)+3\\ & =q_1(x)(x-1{)}^2(x-3)+2x(x-1)+3\\ & =q_1(x)(x-1{)}^2(x-3)+2x^2-2x+3.\end{array}$$This tells us that the remainder when $f(x)$ is divided by $(x-3)(x-1{)}^2$ is $\boxed{2x^2-2x+3}.$