(b) Use implicit differentiation to find \( \frac{\partial z}{\partial y} \) for \[ x e^{2 z}+x=\ln (x)+2 y^{2} z+e \] at \( (x, y, z)=(1,1,1 / 2) \)

We start with the equation \[ x e^{2z}+x=\ln(x)+2y^2 z+e. \] Define \[ F(x,y,z)=x e^{2z}+x-\ln(x)-2y^2 z-e=0. \] We differentiate implicitly with respect to \(y\), treating \(x\) as a constant and writing \(z=z(x,y)\). Differentiate each term: 1. For \(x e^{2z}\), using the chain rule: \[ \frac{\partial}{\partial y}\Bigl(x e^{2z}\Bigr)=x\frac{\partial}{\partial y}\Bigl(e^{2z}\Bigr)=x e^{2z}\cdot\left(2\frac{\partial z}{\partial y}\right)=2x e^{2z}\frac{\partial z}{\partial y}. \] 2. For \(x\) and \(-\ln(x)\), since \(x\) is constant: \[ \frac{\partial}{\partial y}(x)=0,\quad \frac{\partial}{\partial y}\Bigl(-\ln(x)\Bigr)=0. \] 3. For \(-2y^2z\), we use the product rule: \[ \frac{\partial}{\partial y}\Bigl(-2y^2z\Bigr)=-2\left(2yz+y^2\frac{\partial z}{\partial y}\right)=-4yz-2y^2\frac{\partial z}{\partial y}. \] 4. The derivative of the constant \(-e\) is zero. Combining these results, we obtain: \[ 2x e^{2z}\frac{\partial z}{\partial y} -4yz -2y^2 \frac{\partial z}{\partial y} = 0. \] Group the terms with \(\frac{\partial z}{\partial y}\): \[ \left(2x e^{2z} -2y^2\right)\frac{\partial z}{\partial y} - 4yz = 0. \] Solving for \(\frac{\partial z}{\partial y}\), \[ \left(2x e^{2z} -2y^2\right)\frac{\partial z}{\partial y} = 4yz, \] \[ \frac{\partial z}{\partial y} = \frac{4yz}{2x e^{2z}-2y^2}. \] Simplify by canceling the common factor of 2: \[ \frac{\partial z}{\partial y} = \frac{2yz}{x e^{2z}-y^2}. \] Now, evaluate the derivative at \((x,y,z)=(1,1,1/2)\): - The numerator becomes: \[ 2(1)\left(\frac{1}{2}\right)=1. \] - The denominator becomes: \[ 1\cdot e^{2\cdot(1/2)}-1^2=e^1-1=e-1. \] Thus, \[ \frac{\partial z}{\partial y}\bigg|_{(1,1,\frac{1}{2})} = \frac{1}{e-1}. \]