For the function $f(x)=\cos^{-1}\left(\frac{x^2-4x+2}{x^2+3}\right),$ the argument of $\cos^{-1}$ must satisfy $-1 \le \frac{x^2-4x+2}{x^2+3} \le 1.$

First note that $x^2+3>0 \quad \text{for all real }x,$ so we can safely multiply inequalities by $x^2+3$ without changing signs.

Solve the right inequality: $\frac{x^2-4x+2}{x^2+3} \le 1.$ Multiplying by $x^2+3$, $x^2-4x+2 \le x^2+3.$ $-4x+2 \le 3$ $-4x \le 1$ $x \ge -\frac14.$

Solve the left inequality: $\frac{x^2-4x+2}{x^2+3} \ge -1.$ Multiplying by $x^2+3$, $x^2-4x+2 \ge -(x^2+3).$ $x^2-4x+2 \ge -x^2-3$ $2x^2-4x+5 \ge 0.$

Check whether $2x^2-4x+5 \ge 0$ is always true. Its discriminant is $\Delta = (-4)^2-4\cdot 2\cdot 5=16-40=-24<0.$ Since the coefficient of $x^2$ is positive, the quadratic is always positive for all real $x$. Hence this inequality imposes no restriction.

Therefore the only restriction comes from step 3: $x\ge -\frac14.$

So the domain is $\boxed{\left[-\frac14,\infty\right)}.$

Comparing with the options, this is Option B.