Step 1: Find the first three terms in ascending powers of $y$ in the expansion of $(1-y)^{\frac{1}{2}}$. We use the binomial series expansion for $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$. In this case, $x=-y$ and $n=\frac{1}{2}$. The first term is: $$ 1 $$ The second term is $nx$: $$ \left(\frac{1}{2}\right)(-y) = -\frac{1}{2}y $$ The third term is $\frac{n(n-1)}{2!}x^2$: $$ \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}(-y)^2 $$ $$ = \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2 \times 1} (y^2) $$ $$ = \frac{-\frac{1}{4}}{2} y^2 $$ $$ = -\frac{1}{8}y^2 $$ The first three terms in ascending powers of $y$ are $\boxed{1 - \frac{1}{2}y - \frac{1}{8}y^2}$. Step 2: Find the coefficient of $x^{-1}$ in the expansion of $\left((x+1)\left(2x-\frac{1}{x}\right)^3\right)$. First, we expand $\left(2x-\frac{1}{x}\right)^3$ using the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Let $a=2x$ and $b=-\frac{1}{x} = -x^{-1}$. $$ \left(2x-\frac{1}{x}\right)^3 = (2x)^3 + 3(2x)^2\left(-x^{-1}\right) + 3(2x)\left(-x^{-1}\right)^2 + \left(-x^{-1}\right)^3 $$ $$ = 8x^3 + 3(4x^2)(-x^{-1}) + 3(2x)(x^{-2}) + (-x^{-3}) $$ $$ = 8x^3 - 12x^{2-1} + 6x^{1-2} - x^{-3} $$ $$ = 8x^3 - 12x + 6x^{-1} - x^{-3} $$ Now, we need to find the coefficient of $x^{-1}$ in the expansion of $(x+1)\left(8x^3 - 12x + 6x^{-1} - x^{-3}\right)$. We multiply each term in the first bracket by each term in the second bracket and identify terms that result in $x^{-1}$. Terms from $x \times \left(8x^3 - 12x + 6x^{-1} - x^{-3}\right)$: • $x \cdot (8x^3) = 8x^4$ • $x \cdot (-12x) = -12x^2$ • $x \cdot (6x^{-1}) = 6x^0 = 6$ • $x \cdot (-x^{-3}) = -x^{-2}$ None of these terms result in $x^{-1}$. Terms from $1 \times \left(8x^3 - 12x + 6x^{-1} - x^{-3}\right)$: • $1 \cdot (8x^3) = 8x^3$ • $1 \cdot (-12x) = -12x$ • $1 \cdot (6x^{-1}) = 6x^{-1}$ (This term has $x^{-1}$) • $1 \cdot (-x^{-3}) = -x^{-3}$ The only term that results in $x^{-1}$ is $6x^{-1}$. Therefore, the coefficient of $x^{-1}$ is $\boxed{6}$.