f:R→R f((x-y)²)= (f(x))² -2xf(y) +y² then f(x)=?

Setting x = y = 0, we get f(0) = f²(0) which means f(0) = 0 or f(0) = 1

Swapping x and y we get

f[(y-x)²] = f[(x-y)²] = f²(y) - 2y f(x) + x²

Hence we get f²(x) - 2x f(y) + y² = f²(y) - 2y f(x) + x²

Now putting y = 0, we get f²(x) - 2x f(0)  = f²(0) - x²

If f(0) = 0, then f²(x) = x² and we get f(x) = x after discarding f(x) = -x as extraneous

Likewise, if f(0) = 1, then f(x) = x+1 satisfies the given equation