1. **State the problem:** Given that $\left(x+\frac{1}{x}\right)^2=40$, find the value of $x^2+\frac{1}{x^2}$.\n\n2. **Recall the formula:** We know that $\left(a+b\right)^2 = a^2 + 2ab + b^2$. Applying this to $x + \frac{1}{x}$, we get:\n$$\left(x+\frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}.$$\n\n3. **Use the given equation:** Substitute the given value 40 into the equation:\n$$x^2 + 2 + \frac{1}{x^2} = 40.$$\n\n4. **Isolate the desired expression:** We want $x^2 + \frac{1}{x^2}$, so subtract 2 from both sides:\n$$x^2 + \frac{1}{x^2} = 40 - 2 = 38.$$\n\n5. **Final answer:**\n$$\boxed{38}.$$