Answer: $f(x) = x$ for all real $x$.
From the first equation we have by induction that
$$f(x+n) = f(x) + n \quad (1)$$
for every real $x$ and integer $n$. Since $x^4 - x^2 = (x^2 - \frac{1}{2})^2 - \frac{1}{4}$, for every $y \ge -\frac{1}{4}$ there is $x$ such that $y = x^4 - x^2$, implying
$$f(y) = f(x^4 - x^2) = f(x)^4 - f(x)^2 = \left(f(x)^2 - \frac{1}{2}\right)^2 - \frac{1}{4} \ge -\frac{1}{4}.$$

For every $x$ we have $x = \{x\} + [x]$ and $0 \le \{x\} < 1$ (in particular, $\{x\} > -\frac{1}{4}$ and $[x] > x - 1$),
implying
$$f(x) = f(\{x\}) + [x] > -\frac{1}{4} + (x-1) > x - 2. \quad (2)$$
Suppose there is $x$ such that $t = f(x) - x \ne 0$. Then for every integer $n$ we have
$$f(x+n)^4 - f(x+n)^2 = f((x+n)^4 - (x+n)^2),$$
giving
$$(x+n+t)^4 - (x+n+t)^2 > (x+n)^4 - (x+n)^2 - 2,$$
and thus
$$4t(x+n)^3 + 6t^2(x+n)^2 + 2t(2t^2-1)(x+n) + t^4 - t^2 + 2 > 0 \quad (3)$$
for every integer $n$, which is impossible as if $t > 0$, then for $n \to -\infty$ we get a contradiction. If $t < 0$, then for $n \to \infty$ we again get a contradiction. Therefore, $f(x) = x$ for all real $x$ which is easy to check works for both given equations.