1. f(a)≠0 for some a in R
2. f(xf(y))=yf(x) for all x,y in R

My solution is similar to JLo's:

1: proving f(0)=0
a) f(0*f(0))=0*f(0) which means f(0)=0

2: proving f(nonzero)=nonzero
a) assume f(b)=0 for some nonzero b
b) f(a*f(b))=b*f(a) which means f(0)/b=f(a)
c) f(0)/b=0, so f(a)=0
d) this contradicts condition 1, so the assumption must be false

3: proving f(1)=1
a) let f(1)=c
b) f(1*f(1))=1*f(1) which means f(c)=c
c) f(c*f(1))=1*f(c) which means f(c*c)=c
d) f(c*f(c))=c*f(c) which means f(c*c)=c*c
e) since c=c*c and f(nonzero)=nonzero, c=1, thus f(1)=1

4: proving f(-1)=-1
a) let f(-1)=d
b) f(1*f(-1))=-1*f(1) which means f(d)=-1
c) f(-1*f(d))=d*f(-1) which means f(1)=d*d and d=1 or -1
d) f(d*f(d))=d*f(d) which means f(-d)=-d
e) if d=1, then f(-1)=-1, but f(-1)=d=1
f) this leads to a contradiction so d=-1, and f(-1)=-1

5: proving f(-x)=-f(x)
a) f(x*f(-1))=-1*f(x) which means f(-x)=-f(x) since f(-1)=-1