Here is a proof of $d = \frac{1}{2}$. Then you can argue as Francesco did in point two of his answer.

Consider the action: $$ \begin{array}{ccc} \mu_{2}\times\mathbb{A}^{3} & \longrightarrow & \mathbb{A}^{3}\\ (\epsilon,x_{1}, x_{2}, x_3) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}, \epsilon x_3) \end{array} $$ The ring of invariants is given by: $$k[x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2]\cong \frac{k[y_0,y_1,y_2,y_3,y_4,y_5]}{(y_0y_3-y_1^2,y_0y_4-y_1y_2,y_0y_5-y_2^2,y_1y_4-y_2y_3,y_1y_5-y_2y_4,y_3y_5-y_4^2)}$$ The singularity $X=\mathbb{A}^{3}/\mu_{2}$ corresponds to the vertex $v$ of the affine cone $X$ over a Veronese surface $V\subset\mathbb{P}^5$. The differential form $dx_0\wedge dx_1\wedge dx_2$ is a basis of $\bigwedge^3\Omega_{\mathbb{A}^3}$, and $(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}$ is invariant under the action. The form $$\omega = \frac{(dy_0\wedge dy_1\wedge dy_2)^{\otimes 2}}{y_0^3}\in \left(\bigwedge^3\Omega_{k(X)}\right)^{\otimes 2}$$ is a basis of $(\bigwedge^3\Omega_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^3\rightarrow X$ is 'etale on $X\setminus\{v\}$, and $$\pi^{*}\omega = \frac{(4x_0^6 (dx_0\wedge dx_1\wedge dx_2))^{\otimes 2}}{x_{0}^{6}} = 4(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}.$$ Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$, and we have an affine chart isomorphic to $\mathbb{A}^3$ with coordinates $(y_0,s,t)$ where the resolution is given by $(y_0,s,t)\mapsto (y_0,y_0s,t,y_0s^2,y_0st,t^2)$, and the exceptional divisor $E$ over $v$ is given by $\{y_0=0\}$. We have $$f^{*}\omega = \frac{(dy_0\wedge ds\wedge dt)^{\otimes 2}}{y_0}.$$ Therefore, $f^{*}\omega$ has a pole along $E$, and we may write $K_Y = f^{*}K_X+\frac{1}{2}E$.

Just to give you an idea of the computations which are involved here is the simpler example of a quadric cone.

Let us consider the action: $$ \begin{array}{ccc} \mu_{2}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\ (\epsilon,x_{1}, x_{2}) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}) \end{array} $$ The ring of invariants is given by: $$k[x_0^2,x_0x_1,x_1^2]\cong k[y_0,y_1,y_2]/(y_0y_2-y_1^2)$$ and we see that the singularity $X=\mathbb{A}^{2}/\mu_{2}$ corresponds to the vertex $v$ of the affine cone $$X=Spec(k[x_0^2,x_0x_1,x_1^2]\cong k[y_0,y_1,y_2]/(y_0y_2-y_1^2))$$ that is the vertex of a quadric cone $Q\subset\mathbb{P}^2$ or equivalently the singularity $\frac{1}{2}(1,1)$ of the weighted projective plane $\mathbb{P}(1,1,2)$. Now, $dx_0\wedge dx_1$ is a basis of $\bigwedge^2\Omega_{\mathbb{A}^2}$, and $(dx_0\wedge dx_1)^{\otimes 2}$ is invariant under the action. The form $$\omega = \frac{(dy_0\wedge dy_1)^{\otimes 2}}{y_0^2}\in (\bigwedge^2\Omega_{k(X)})^{\otimes 2}$$ is a basis of $(\bigwedge^2\Omega_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^2\rightarrow X$ is 'etale on $X\setminus\{v\}$, and $\pi^{*}\omega = 4(dx_0\wedge dx_1)^{\otimes 2}$.\ Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$. If $[\lambda_0:\lambda_1:\lambda_2]$ are homogeneous coordinates on $\mathbb{P}^2$ then the equations of $Y$ in $\mathbb{A}^3\times\mathbb{P}^2$ are: $$ \left\lbrace\begin{array}{l} y_0\lambda_1-y_1\lambda_0=0, \\ y_0\lambda_2-y_2\lambda_0=0, \\ y_1\lambda_2-y_2\lambda_1=0, \\ y_0y_2-y_1^2. \end{array}\right. $$ Therefore, $y_1 = \frac{\lambda_1}{\lambda_0}y_0$, and $\frac{\lambda_2}{\lambda_1}=\frac{\lambda_1}{\lambda_0}$ yields $y_2 = \frac{\lambda_1}{\lambda_0}y_1 = (\frac{\lambda_1}{\lambda_0})^2y_0$. Then, in $Y$ we have an affine chart isomorphic to $\mathbb{A}^2$ with coordinates $(y_0,t)$ where the resolution is given by $(y_0,t)\mapsto (y_0,y_0t,y_0t^2)$, with $t = \frac{\lambda_1}{\lambda_0}$, and the exceptional divisor $E$ over $v$ is given by $\{y_0=0\}$. We have $$f^{*}\omega = (dy_0\wedge dt)^{\otimes 2}.$$ Therefore, $f^{*}\omega$ has neither a pole nor a zero along $E$, and we may write $K_Y = f^{*}K_X$.

Here is a proof of $d = \frac{1}{2}$. Then you can argue as Francesco did in point two of his answer.

Consider the action: $$ \begin{array}{ccc} \mu_{2}\times\mathbb{A}^{3} & \longrightarrow & \mathbb{A}^{3}\\ (\epsilon,x_{1}, x_{2}, x_3) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}, \epsilon x_3) \end{array} $$ The ring of invariants is given by: $$k[x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2]\cong \frac{k[y_0,y_1,y_2,y_3,y_4,y_5]}{(y_0y_3-y_1^2,y_0y_4-y_1y_2,y_0y_5-y_2^2,y_1y_4-y_2y_3,y_1y_5-y_2y_4,y_3y_5-y_4^2)}$$ The singularity $X=\mathbb{A}^{3}/\mu_{2}$ corresponds to the vertex $v$ of the affine cone $X$ over a Veronese surface $V\subset\mathbb{P}^5$. The differential form $dx_0\wedge dx_1\wedge dx_2$ is a basis of $\bigwedge^3\Omega_{\mathbb{A}^3}$, and $(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}$ is invariant under the action. The form $$\omega = \frac{(dy_0\wedge dy_1\wedge dy_2)^{\otimes 2}}{y_0^3}\in \left(\bigwedge^3\Omega_{k(X)}\right)^{\otimes 2}$$ is a basis of $(\bigwedge^3\Omega_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^3\rightarrow X$ is 'etale on $X\setminus\{v\}$, and $$\pi^{*}\omega = \frac{(4x_0^6 (dx_0\wedge dx_1\wedge dx_2))^{\otimes 2}}{x_{0}^{6}} = 4(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}.$$ Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$, and we have an affine chart isomorphic to $\mathbb{A}^3$ with coordinates $(y_0,s,t)$ where the resolution is given by $(y_0,s,t)\mapsto (y_0,y_0s,y_0t,y_0s^2,y_0st,y_0t^2)$, and the exceptional divisor $E$ over $v$ is given by $\{y_0=0\}$. We have $$f^{*}\omega = y_0(dy_0\wedge ds\wedge dt)^{\otimes 2}.$$ Therefore, $f^{*}\omega$ has a zero along $E$, and we may write $K_Y = f^{*}K_X+\frac{1}{2}E$.

Just to give you an idea of the computations which are involved here is the simpler example of a quadric cone.

Let us consider the action: $$ \begin{array}{ccc} \mu_{2}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\ (\epsilon,x_{1}, x_{2}) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}) \end{array} $$ The ring of invariants is given by: $$k[x_0^2,x_0x_1,x_1^2]\cong k[y_0,y_1,y_2]/(y_0y_2-y_1^2)$$ and we see that the singularity $X=\mathbb{A}^{2}/\mu_{2}$ corresponds to the vertex $v$ of the affine cone $$X=Spec(k[x_0^2,x_0x_1,x_1^2]\cong k[y_0,y_1,y_2]/(y_0y_2-y_1^2))$$ that is the vertex of a quadric cone $Q\subset\mathbb{P}^2$ or equivalently the singularity $\frac{1}{2}(1,1)$ of the weighted projective plane $\mathbb{P}(1,1,2)$. Now, $dx_0\wedge dx_1$ is a basis of $\bigwedge^2\Omega_{\mathbb{A}^2}$, and $(dx_0\wedge dx_1)^{\otimes 2}$ is invariant under the action. The form $$\omega = \frac{(dy_0\wedge dy_1)^{\otimes 2}}{y_0^2}\in \left(\bigwedge^2\Omega_{k(X)}\right)^{\otimes 2}$$ is a basis of $(\bigwedge^2\Omega_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^2\rightarrow X$ is 'etale on $X\setminus\{v\}$, and $\pi^{*}\omega = 4(dx_0\wedge dx_1)^{\otimes 2}$.\ Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$. If $[\lambda_0:\lambda_1:\lambda_2]$ are homogeneous coordinates on $\mathbb{P}^2$ then the equations of $Y$ in $\mathbb{A}^3\times\mathbb{P}^2$ are: $$ \left\lbrace\begin{array}{l} y_0\lambda_1-y_1\lambda_0=0, \\ y_0\lambda_2-y_2\lambda_0=0, \\ y_1\lambda_2-y_2\lambda_1=0, \\ y_0y_2-y_1^2. \end{array}\right. $$ Therefore, $y_1 = \frac{\lambda_1}{\lambda_0}y_0$, and $\frac{\lambda_2}{\lambda_1}=\frac{\lambda_1}{\lambda_0}$ yields $y_2 = \frac{\lambda_1}{\lambda_0}y_1 = (\frac{\lambda_1}{\lambda_0})^2y_0$. Then, in $Y$ we have an affine chart isomorphic to $\mathbb{A}^2$ with coordinates $(y_0,t)$ where the resolution is given by $(y_0,t)\mapsto (y_0,y_0t,y_0t^2)$, with $t = \frac{\lambda_1}{\lambda_0}$, and the exceptional divisor $E$ over $v$ is given by $\{y_0=0\}$. We have $$f^{*}\omega = (dy_0\wedge dt)^{\otimes 2}.$$ Therefore, $f^{*}\omega$ has neither a pole nor a zero along $E$, and we may write $K_Y = f^{*}K_X$.

Here is a proof of $d = \frac{1}{2}$. Then you can argue as Francesco did in point two of his answer.

Consider the action: $$ \begin{array}{ccc} \mu_{2}\times\mathbb{A}^{3} & \longrightarrow & \mathbb{A}^{3}\\ (\epsilon,x_{1}, x_{2}, x_3) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}, \epsilon x_3) \end{array} $$ The ring of invariants is given by: $$k[x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2]\cong \frac{k[y_0,y_1,y_2,y_3,y_4,y_5]}{(y_0y_3-y_1^2,y_0y_4-y_1y_2,y_0y_5-y_2^2,y_1y_4-y_2y_3,y_1y_5-y_2y_4,y_3y_5-y_4^2)}$$ The singularity $X=\mathbb{A}^{3}/\mu_{2}$ corresponds to the vertex $v$ of the affine cone $X$ over a Veronese surface $V\subset\mathbb{P}^5$. The differential form $dx_0\wedge dx_1\wedge dx_2$ is a basis of $\bigwedge^3\Omega_{\mathbb{A}^3}$, and $(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}$ is invariant under the action. The form $$\omega = \frac{(dy_0\wedge dy_1\wedge dy_2)^{\otimes 2}}{y_0^3}\in \left(\bigwedge^3\Omega_{k(X)}\right)^{\otimes 2}$$ is a basis of $(\bigwedge^3\Omega_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^3\rightarrow X$ is 'etale on $X\setminus\{v\}$, and $$\pi^{*}\omega = \frac{(4x_0^6 (dx_0\wedge dx_1\wedge dx_2))^{\otimes 2}}{x_{0}^{6}} = 4(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}.$$ Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$, and we have an affine chart isomorphic to $\mathbb{A}^3$ with coordinates $(y_0,s,t)$ where the resolution is given by $(y_0,s,t)\mapsto (y_0,y_0s,t,y_0s^2,y_0st,t^2)$, and the exceptional divisor $E$ over $v$ is given by $\{y_0=0\}$. We have $$f^{*}\omega = \frac{(dy_0\wedge ds\wedge dt)^{\otimes 2}}{y_0}.$$ Therefore, $f^{*}\omega$ has a pole along $E$, and we may write $K_Y = f^{*}K_X+\frac{1}{2}E$.

Here is a proof of $d = \frac{1}{2}$. Then you can argue as Francesco did in point two of his answer.

Consider the action: $$ \begin{array}{ccc} \mu_{2}\times\mathbb{A}^{3} & \longrightarrow & \mathbb{A}^{3}\\ (\epsilon,x_{1}, x_{2}, x_3) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}, \epsilon x_3) \end{array} $$ The ring of invariants is given by: $$k[x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2]\cong \frac{k[y_0,y_1,y_2,y_3,y_4,y_5]}{(y_0y_3-y_1^2,y_0y_4-y_1y_2,y_0y_5-y_2^2,y_1y_4-y_2y_3,y_1y_5-y_2y_4,y_3y_5-y_4^2)}$$ The singularity $X=\mathbb{A}^{3}/\mu_{2}$ corresponds to the vertex $v$ of the affine cone $X$ over a Veronese surface $V\subset\mathbb{P}^5$. The differential form $dx_0\wedge dx_1\wedge dx_2$ is a basis of $\bigwedge^3\Omega_{\mathbb{A}^3}$, and $(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}$ is invariant under the action. The form $$\omega = \frac{(dy_0\wedge dy_1\wedge dy_2)^{\otimes 2}}{y_0^3}\in \left(\bigwedge^3\Omega_{k(X)}\right)^{\otimes 2}$$ is a basis of $(\bigwedge^3\Omega_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^3\rightarrow X$ is 'etale on $X\setminus\{v\}$, and $$\pi^{*}\omega = \frac{(4x_0^6 (dx_0\wedge dx_1\wedge dx_2))^{\otimes 2}}{x_{0}^{6}} = 4(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}.$$ Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$, and we have an affine chart isomorphic to $\mathbb{A}^3$ with coordinates $(y_0,s,t)$ where the resolution is given by $(y_0,s,t)\mapsto (y_0,y_0s,t,y_0s^2,y_0st,t^2)$, and the exceptional divisor $E$ over $v$ is given by $\{y_0=0\}$. We have $$f^{*}\omega = \frac{(dy_0\wedge ds\wedge dt)^{\otimes 2}}{y_0}.$$ Therefore, $f^{*}\omega$ has a pole along $E$, and we may write $K_Y = f^{*}K_X+\frac{1}{2}E$.

Just to give you an idea of the computations which are involved here is the simpler example of a quadric cone.

Let us consider the action: $$ \begin{array}{ccc} \mu_{2}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\ (\epsilon,x_{1}, x_{2}) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}) \end{array} $$ The ring of invariants is given by: $$k[x_0^2,x_0x_1,x_1^2]\cong k[y_0,y_1,y_2]/(y_0y_2-y_1^2)$$ and we see that the singularity $X=\mathbb{A}^{2}/\mu_{2}$ corresponds to the vertex $v$ of the affine cone $$X=Spec(k[x_0^2,x_0x_1,x_1^2]\cong k[y_0,y_1,y_2]/(y_0y_2-y_1^2))$$ that is the vertex of a quadric cone $Q\subset\mathbb{P}^2$ or equivalently the singularity $\frac{1}{2}(1,1)$ of the weighted projective plane $\mathbb{P}(1,1,2)$. Now, $dx_0\wedge dx_1$ is a basis of $\bigwedge^2\Omega_{\mathbb{A}^2}$, and $(dx_0\wedge dx_1)^{\otimes 2}$ is invariant under the action. The form $$\omega = \frac{(dy_0\wedge dy_1)^{\otimes 2}}{y_0^2}\in (\bigwedge^2\Omega_{k(X)})^{\otimes 2}$$ is a basis of $(\bigwedge^2\Omega_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^2\rightarrow X$ is 'etale on $X\setminus\{v\}$, and $\pi^{*}\omega = 4(dx_0\wedge dx_1)^{\otimes 2}$.\ Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$. If $[\lambda_0:\lambda_1:\lambda_2]$ are homogeneous coordinates on $\mathbb{P}^2$ then the equations of $Y$ in $\mathbb{A}^3\times\mathbb{P}^2$ are: $$ \left\lbrace\begin{array}{l} y_0\lambda_1-y_1\lambda_0=0, \\ y_0\lambda_2-y_2\lambda_0=0, \\ y_1\lambda_2-y_2\lambda_1=0, \\ y_0y_2-y_1^2. \end{array}\right. $$ Therefore, $y_1 = \frac{\lambda_1}{\lambda_0}y_0$, and $\frac{\lambda_2}{\lambda_1}=\frac{\lambda_1}{\lambda_0}$ yields $y_2 = \frac{\lambda_1}{\lambda_0}y_1 = (\frac{\lambda_1}{\lambda_0})^2y_0$. Then, in $Y$ we have an affine chart isomorphic to $\mathbb{A}^2$ with coordinates $(y_0,t)$ where the resolution is given by $(y_0,t)\mapsto (y_0,y_0t,y_0t^2)$, with $t = \frac{\lambda_1}{\lambda_0}$, and the exceptional divisor $E$ over $v$ is given by $\{y_0=0\}$. We have $$f^{*}\omega = (dy_0\wedge dt)^{\otimes 2}.$$ Therefore, $f^{*}\omega$ has neither a pole nor a zero along $E$, and we may write $K_Y = f^{*}K_X$.