Determine all possible (finite or infinite) values of
\[\lim_{x \to -\infty} f(x)-\lim_{x \to \infty} f(x),\]
if $f:\mathbb{R} \to \mathbb{R}$ is a strictly decreasing continuous function satisfying
\[f(f(x))^4-f(f(x))+f(x)=1\]
for all $x \in \mathbb{R}$.
1. **Define Limits and Function Properties:**
   Let 
   \[
   \lim_{x \to -\infty} f(x) = M \quad \text{and} \quad \lim_{x \to \infty} f(x) = m.
   \]
   Since \( f \) is strictly decreasing and continuous, it follows that \( m < f(x) < M \) for all \( x \in \mathbb{R} \). This implies that \( f \) maps \(\mathbb{R}\) to the interval \((m, M)\) bijectively.

2. **Transform the Given Functional Equation:**
   The given functional equation is:
   \[
   f(f(x))^4 - f(f(x)) + f(x) = 1.
   \]
   Let \( t = f(x) \). Then the equation becomes:
   \[
   f(t)^4 - f(t) + t = 1.
   \]
   Define a new function \( g(u) = -u^4 + u + 1 \). The equation can be rewritten as:
   \[
   t = g(f(t)).
   \]

3. **Analyze the Function \( g \):**
   The derivative of \( g \) is:
   \[
   g'(u) = -4u^3 + 1.
   \]
   Setting \( g'(u) = 0 \) gives:
   \[
   -4u^3 + 1 = 0 \implies u_0 = \frac{1}{\sqrt[3]{4}} \approx 0.63.
   \]
   Therefore, \( g(u) \) is strictly increasing on \((-\infty, u_0]\) and strictly decreasing on \([u_0, +\infty)\).

4. **Behavior of \( f \) and \( g \circ f \):**
   Since \( f \) is strictly decreasing, \( f(t) \) is strictly decreasing on \((m, M)\). For \( (g \circ f)(t) = t \) to hold, \( f(t) \) must be greater than or equal to \( u_0 \) for all \( t \in (m, M) \). This implies \( f(t) \ge u_0 \) for all \( t \in (m, M) \), and thus \( g(u) \) is strictly decreasing.

5. **Bounds on \( M \) and \( m \):**
   Since \( g(u) \le g(u_0) = 1 + \frac{3}{4\sqrt[3]{4}} \approx 1.47 \) for all \( u \in (A, B) \), it follows that:
   \[
   t = f(x) \le 1 + \frac{3}{4\sqrt[3]{4}} \quad \forall x \in \mathbb{R}.
   \]
   Therefore,
   \[
   M = \lim_{x \to -\infty} f(x) \le 1 + \frac{3}{4\sqrt[3]{4}}.
   \]

6. **Determine \( m \):**
   Using the equation \( f(f(x))^4 - f(f(x)) + f(x) = 1 \), we have:
   \[
   f(x) = (g \circ f \circ f)(x).
   \]
   Taking the limit as \( x \to +\infty \), we get:
   \[
   m = g(f(m)).
   \]
   Since \( f(m) \le M \le 1 + \frac{3}{4\sqrt[3]{4}} \) and \( g \) is strictly decreasing, it follows that:
   \[
   m \ge g(M) \ge g\left(1 + \frac{3}{4\sqrt[3]{4}}\right) = \frac{37}{64} - \frac{27}{16\sqrt[3]{2}} - \frac{2385}{1024\sqrt[3]{4}} \approx -2.23.
   \]
   Thus, \( m \) is finite and \( f(m) < M \), implying \( m > g(M) \ge 1 + \frac{3}{4\sqrt[3]{4}} \).

7. **Range of \( M - m \):**
   The difference \( M - m \) is positive and bounded by:
   \[
   0 < M - m < M - g(M) \le \max(M) - g(\max(M)) = \frac{27}{64} + \frac{27}{16\sqrt[3]{2}} + \frac{3153}{1024\sqrt[3]{4}} \approx 3.70.
   \]

8. **Conclusion:**
   The set of all possible values of \( \lim_{x \to -\infty} f(x) - \lim_{x \to \infty} f(x) \) is the interval:
   \[
   S = \left(0, \frac{27}{64} + \frac{27}{16\sqrt[3]{2}} + \frac{3153}{1024\sqrt[3]{4}}\right).
   \]

The final answer is \( \boxed{ \left(0, \frac{27}{64} + \frac{27}{16\sqrt[3]{2}} + \frac{3153}{1024\sqrt[3]{4}}\right) } \).