(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 $$ (x, y, z)=(1,1,1 / 2) $$

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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,
$$
\frac{\partial z}{\partial y}\bigg|_{(1,1,\frac{1}{2})} = \frac{1}{e-1}.
$$
At the point $$(1,1,1/2)$$, the derivative $$\frac{\partial z}{\partial y}$$ is $$\frac{1}{e-1}$$.

(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 \( (x, y, z)=(1,1,1 / 2) \)

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