For any non-zero real number x, let f(x) + 2f 1/x = 3x. Then, the sum of all possible values of x for which f(x) = 3 is:

The correct option is -3.

To find the sum of all possible values of $x$ for which $f\left(x\right)=3$, we first need to determine the function $f\left(x\right)$ from the given functional equation:

$$f\left(x\right)+2f\left(\frac{1}{x}\right)=3x$$

Let this be Equation (1). Since this relation holds for any non-zero real number $x$, we can substitute $x$ with $\frac{1}{x}$ in Equation (1) to obtain:

$$f\left(\frac{1}{x}\right)+2f\left(x\right)=3\left(\frac{1}{x}\right)$$

Rearranging the terms on the left-hand side, we get Equation (2):

$$2f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{3}{x}$$

Now we have a system of two linear equations in terms of $f\left(x\right)$ and $f\left(\frac{1}{x}\right)$. To solve for $f\left(x\right)$, we can eliminate the term $f\left(\frac{1}{x}\right)$. We multiply Equation (2) by 2:

$$4f\left(x\right)+2f\left(\frac{1}{x}\right)=\frac{6}{x}$$

Let this be Equation (3). Next, we subtract Equation (1) from Equation (3):

$$\left[4f\right(x)+2f\left(\frac{1}{x}\right)]-\left[f\right(x)+2f\left(\frac{1}{x}\right)]=\frac{6}{x}-3x$$

Simplifying the equation yields:

$$3f\left(x\right)=\frac{6}{x}-3x$$

Dividing by 3, we obtain the function $f\left(x\right)$:

$$f\left(x\right)=\frac{2}{x}-x$$

We want to find the values of $x$ for which $f\left(x\right)=3$. Setting the function equal to 3 gives:

$$\frac{2}{x}-x=3$$

To eliminate the denominator, we multiply both sides of the equation by $x$ (since $x\ne 0$):

$$2-x^2=3x$$

Rearranging the terms into standard quadratic form $a{x}^2+bx+c=0$:

$$x^2+3x-2=0$$

We check the discriminant $D$ of this quadratic equation to ensure it has real roots:

$$D=b^2-4ac=3^2-4\left(1\right)(-2)=9+8=17$$

Since the discriminant $D>0$, there are two distinct real roots. Furthermore, neither root is zero because substituting $x=0$ into the equation gives $-2\ne 0$. Therefore, both values are valid non-zero real inputs for our function.

By Vieta's formulas, the sum of the roots of the quadratic equation $a{x}^2+bx+c=0$ is given by:

$$\text{Sum of roots}=-\frac{b}{a}$$

Using the coefficients $a=1$ and $b=3$:

$$\text{Sum of roots}=-\frac{3}{1}=-3$$

Thus, the sum of all possible values of $x$ for which $f\left(x\right)=3$ is $-3$.