Answer: \((x,y)=(-2,\frac12)\) or \((\frac14,\frac18)\).

Start with

\(\dfrac{x}{2y}-\dfrac{4y}{x}=-1.\)

Since \(x\) and \(y\) appear in denominators, any solution must have \(x\ne0\) and \(y\ne0\). Multiply through by \(2xy\):

\(x^2-8y^2=-2xy.\)

Bring all terms to one side:

\(x^2+2xy-8y^2=0.\)

Factorise the quadratic expression:

\((x+4y)(x-2y)=0.\)

So either

\(x=-4y\quad\text{or}\quad x=2y.\)

Using the second equation \(x=1-6y\), first take \(x=-4y\):

\(-4y=1-6y.\)

Thus \(2y=1\), so \(y=\dfrac12\), and then \(x=-2\).

Now take \(x=2y\):

\(2y=1-6y.\)

Thus \(8y=1\), so \(y=\dfrac18\), and then \(x=\dfrac14\).

Therefore

\((x,y)=\left(-2,\dfrac12\right)\quad\text{or}\quad \left(\dfrac14,\dfrac18\right).\)