Take logs of each side:
½x2log(x-1/x)=-(1+x)log(2),
½x2(log(x2-1)-log(x))+(1+x)log(2)=0.
Let f(x)=½x2(log(x2-1)-log(x))+(1+x)log(2). Define log(x) as ln(x), because any log base is valid and because ln(x), etc., is easy to differentiate.
f'(x)=xln(x-1/x)+½x2(2x/(x2-1)-1/x)+ln(2),
f'(x)=xln(x-1/x)+x3/(x2-1)-x/2+ln(2).
Newton's iterative Method:
xn+1=xn-f(xn)/f'(xn). We need to choose x0 to start the process. We must avoid x0=0 and x0-1/x0≤0 which would cause anomalies in the expressions. Choose x0=1.5.
x1=1.00913, x2=1.02155..., ..., x=1.03681291616 after a few more iterations.
However, this is not the only solution, so we need to set x0 to a different value nearer the other solution. A rough sketch shows that x0=-0.75 is a promising start value. After a few iterations, x=-0.75871567976.
So there are two solutions as calculated above.