Let $f$ be a differentiable function satisfying $f(x)+f(y)+f(z)+f(x) f(y) f(z)=14$ for all $x, y, z \in R$ Then,

$f^{\prime}(x)=0$ for all $x \in R$

We have,

$f(x)+f(y)+f(z)+f(x) f(y) f(z)=14$ for all $x, y, z \in R$            ......(i)

Putting x = y = z = 0, we get

$3 f(0)+\{f(0)\}^3=14$

$\Rightarrow \{f(0)\}^3+3 f(0)-14=0$

$\Rightarrow f(0)=2$

Now, putting $y=z=x$ in (i), we get

$3 f(x)+\{f(x)\}^3=14$ for all $x \in R$

Differentiating with respect to $x$, we get

$3 f^{\prime}(x)+3\{f(x)\}^2 f^{\prime}(x)=0$ for all $x \in R$

$\Rightarrow \left\{\{f(x)\}^2+1\right\} f^{\prime}(x)=0$ for all $x \in R$

$\Rightarrow f^{\prime}(x)=0$ for all $x \in R$