derp, philosophy is for boring faggots, let's solve equations!

The presence of two variables x and y within the function boggled me at the start, but this function f(x) still has a single-variable input and output. So I tried to sketch the graph in the 2D coordinate plane and then guess which function would fit. There's probably a more elegant way of solving this.

Let x = y = 1
By the first given,
f(1*1) = 1f(1) + 1f(1)
f(1) = 2f(1)
f(1) = 0

Let x = y = -1
By the first given,
f(1) = -1f(-1) - 1f(-1)
f(1) = -2f(-1)
0 = -2f(-1)
f(-1) = 0

>>307

Let x = y = 0
By the first given,
f(0) = 0

Let x = y
By the first given,
f(x^2) = xf(x) + xf(x)
f(x^2) = 2xf(x) [I think that, in this case, f(x^n) = (x^n)'f(x) for all real n]
Differentiate both sides with respect to x:
2xf'(x^2) = 2f(x) + 2xf'(x)
xf'(x^2) = f(x) + xf'(x)
Now let x = -1
-f'(1) = f(-1) - f'(-1)
-4 = 0 - f'(-1)
f'(-1) = 4

>>308

So now we know
f(1) = f(0) = f(-1) = 0
f'(1) = f'(-1) = 4
And the graph must look something like this:

>>309

But I could only guess at the function's end behavior. Knowing that f(x) was fairly simple, I tried using cubic then sine functions to satisfy the first given condition.

Then I tried f(x) = xsinx

f(xy) = xysin(xy)
x*f(y) + y*f(x) = xysiny + yxsinx

If only sinx + siny = sinxy.... but with logs, this would work. I tried f(x) = xlnx

f(xy) = xyln(xy) = xy(lnx + lny) = xylnx + xylny = y(xlnx) + x(ylny) = y*f(x) + x*f(y)

And just make it f(x) = 4xlnx to satisfy f'(1) =4.

f(x) = 4xlnx
is one solution; maybe there are more.

>>310

Notes:
1. A logarithm of any base can be used: f(x) = 4(lnb)xlog"b"x where log"b"x is the logarithm base b of x and where b is real and positive.
2. In my work, I obtained graph points for x = 0 and -1, but ultimately 4xlnx is undefined when x = 0 or x < 0. f(x) = 4xln|x| would have been a bit closer to what I was imagining, although f(x) = 4xlnx is all you need to solve the problem.

>>311

very nice, you win 2 points