The differentiable function $F:\mathbb{R}\to\mathbb{R}$ satisfies $F(0)=-1$ and \[\dfrac{d}{dx}F(x)=\sin (\sin (\sin (\sin(x))))\cdot \cos( \sin (\sin (x))) \cdot \cos (\sin(x))\cdot\cos(x).\] Find $F(x)$ as a function of $x$.
1. Given the differential equation:
   \[
   \frac{d}{dx}F(x) = \sin (\sin (\sin (\sin(x)))) \cdot \cos(\sin (\sin (x))) \cdot \cos (\sin(x)) \cdot \cos(x)
   \]
   We need to find the function \( F(x) \).

2. We start by recognizing that the right-hand side of the differential equation can be written as a product of derivatives. Notice that:
   \[
   \frac{d}{dx} (\sin (\sin (\sin (x)))) = \cos (\sin (\sin (x))) \cdot \cos (\sin (x)) \cdot \cos (x)
   \]
   Therefore, we can rewrite the differential equation as:
   \[
   \frac{d}{dx}F(x) = \sin (\sin (\sin (\sin(x)))) \cdot \frac{d}{dx} (\sin (\sin (\sin (x))))
   \]

3. This suggests that we can use the chain rule to integrate both sides. Let \( u = \sin (\sin (\sin (x))) \). Then:
   \[
   \frac{d}{dx} u = \cos (\sin (\sin (x))) \cdot \cos (\sin (x)) \cdot \cos (x)
   \]
   Hence, the differential equation becomes:
   \[
   \frac{d}{dx}F(x) = \sin (u) \cdot \frac{d}{dx} u
   \]

4. Integrating both sides with respect to \( x \):
   \[
   \int \frac{d}{dx}F(x) \, dx = \int \sin (u) \cdot \frac{d}{dx} u \, dx
   \]
   The right-hand side simplifies to:
   \[
   F(x) = \int \sin (u) \, du
   \]

5. The integral of \( \sin (u) \) is \( -\cos (u) \), so:
   \[
   F(x) = -\cos (u) + C
   \]
   Substituting back \( u = \sin (\sin (\sin (x))) \):
   \[
   F(x) = -\cos (\sin (\sin (\sin (x)))) + C
   \]

6. We are given the initial condition \( F(0) = -1 \). Substituting \( x = 0 \):
   \[
   F(0) = -\cos (\sin (\sin (\sin (0)))) + C = -\cos (\sin (\sin (0))) + C = -\cos (\sin (0)) + C = -\cos (0) + C = -1 + C
   \]
   Since \( F(0) = -1 \), we have:
   \[
   -1 + C = -1 \implies C = 0
   \]

7. Therefore, the function \( F(x) \) is:
   \[
   F(x) = -\cos (\sin (\sin (\sin (x))))
   \]

The final answer is \(\boxed{F(x) = -\cos (\sin (\sin (\sin (x))))}\).