Determine all functions $f:\mathbb{R}\to \mathbb{R}$ such that, for any real numbers $x$ and $y$,

Step 1 (Initial substitutions): Let $P(x,y)$ denote the assertion $f(f(x)f(y))+f(x+y)=f(xy)$.

$P(0,0)$: $f(f(0{)}^2)+f(0)=f(0)$, so $f(f(0{)}^2)=0$. Let $c=f(0)$, so $f(c^2)=0$.

$P(x,0)$: $f(f(x)\cdot c)+f(x)=f(0)=c$, so $f(cf(x))=c-f(x)$.

Step 2 (Case $c=0$): If $c=f(0)=0$, then $f(0)=0$ and from $P(x,0)$: $f(0)+f(x)=f(0)$, giving $f(x)=0$ for all $x$. Check: $0+0=0$. ✓

Step 3 (Case $c\ne 0$): From $f(cf(x))=c-f(x)$, the function $g(x)=c-f(x)$ satisfies $g(cf(x))=f(x)$ and $f(cg(x))=f(c(c-f(x)))$.

$P(x,1)$: $f(f(x)f(1))+f(x+1)=f(x)$.

Let $f(1)=a$. Then $f(af(x))+f(x+1)=f(x)$.

Step 4 (Trying $f(x)=x-1$): Check: $f(f(x)f(y))+f(x+y)=f((x-1)(y-1))+(x+y-1)=(x-1)(y-1)-1+x+y-1=xy-x-y+1-1+x+y-1=xy-1=f(xy)$. ✓

Step 5 (Trying $f(x)=1-x$): Check: $f(f(x)f(y))+f(x+y)=f((1-x)(1-y))+(1-x-y)=1-(1-x)(1-y)+1-x-y=1-1+x+y-xy+1-x-y=1-xy=f(xy)$. ✓

Step 6 (Uniqueness): Through careful substitution analysis (setting $y=1$, $y=x$, etc.) and using the relation $f(cf(x))=c-f(x)$, one can show these are the only solutions. The key is proving $f$ is injective (when $c\ne 0$) using the functional equation, then deducing linearity.

Answer: $f(x)=0$, $f(x)=x-1$, or $f(x)=1-x$.