First, the minimum value of $x$ is $b$ such that $f(x) - b$ has a (double) root. (That is, the amount you must shift the graph of $f$ down so that it meets the $x$-axis once.) So we are looking for a $b$ so that $f(x) - b = x^4 + \frac{1}{x^2} - b = 0$ has a (double) root. Observe$$ x^4 + \frac{1}{x^2} - b = \frac{x^6 - b x^2 + 1}{x^2} $$has a root exactly when its numerator does. So now we just need to know when that cubic in $x^2$ has a double root.
The discriminant of $(x^2)^3 - b(x^2) + 1$ is $$ -4(-b)^3 - 27 \cdot (1)^2 = 4b^3 - 27 \text{.} $$The discriminant is zero if and only if the polynomial has a double root. Taking $b = \sqrt[3]{27/4} = \frac{3}{2^{2/3}}$ is the only choice that makes the discriminant zero, so the minimum value of $f$ is this $b$.
