Be careful of the constraint!

Substituting $y = 1-x$ we get $x-x^2+(x-x^2)^{-1}$ .

Differentiating this with respect to $x$ we get

$1-2x -(1-2x)(x-x^2)^{-2})$ .

$(1-2x)(1-(x-x^2)^{-2})$

When we set this equation equal to zero and solve for $x$ , we can find the minimum.

Either $1-2x = 0$ or $(1-(x-x^2)^{-2}) = 0, (x-x^2)^{-2}) = 1$

From the first equation we can see that $x = \frac{1}{2}$ and therefore $y = \frac{1}{2}$ .

$\frac{1}{4} + \frac{1}{\frac{1}{4}} = \frac{17}{4}$

From the second equation, we can rearrange to see that:

$\frac{1}{(x-x^2)^2} = 1$

$(x-x^2)^2 = 1$

$x^4 -2x^3 + x^2 -1 = 0$ Which has two solutions outside of the range of $0 < x < 1$ and therefore either $x$ or $y$ are negative. (Not allowed)

Therefore the answer is $\boxed{\frac{17}{4}}$

Because of the constraint $x+y=1$ , we cannot apply the AM-GM inequality directly because we would get $xy=1$ which has no real solutions. Well, we may suspect that the minimum value occurs at $x=y=\frac{1}{2}$ , if that's the case we can do a little trick, we write $xy+\frac{1}{xy}$ as follows:

$xy+\frac{1}{16xy}+\frac{15}{16xy}$ and apply the AM-GM inequality because now, $xy=\frac{1}{16xy}$ has real solution!

$xy+\frac{1}{16xy}+\frac{15}{16xy}\ge2\sqrt{(xy)(\frac{1}{xy})}+\frac{15}{16xy}=2+\frac{15}{16xy}$ , also we know $xy=\frac{1}{16xy}$ , then $x=y=\frac{1}{2}$ .

$\Rightarrow xy+\frac{1}{xy}=xy+\frac{1}{16xy}+\frac{15}{16xy}\ge\frac{1}{2}+\frac{15}{16xy}=\frac{1}{2}+\frac{15}{4}=\frac{17}{4}$

There are still many ways you can solve this problem, namely by using calculus, taking the derivative of the function after substituting $y=1-x$ or if you like partial derivatives, you can also consider using Lagrange Multipliers to destroy the problem.

Minimum occurs when $x=y=\dfrac{1}{2}$ . The minimum value is $(\dfrac{1}{2})^2+2^2$ or $\boxed {\dfrac{17}{4}}$