If $$e^y + xy = e$$, the ordered pair $$\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)$$ at x = 0 is equal to

We are given the implicit relation $$e^{y}+x\,y=e.$$

First we find the value of $$y$$ at the point where $$x=0.$$ Substituting $$x=0$$ gives

$$e^{y}+0\cdot y=e\;\;\Longrightarrow\;\;e^{y}=e\;\;\Longrightarrow\;\;y=1.$$

So, at $$x=0$$ we have the point $$(0,1).$$

Now we need the first derivative $$\dfrac{dy}{dx}.$$ We differentiate the original equation with respect to $$x$$. Using the chain rule for $$e^{y}$$ and the product rule for $$x\,y$$ we have

$$\frac{d}{dx}(e^{y})+\frac{d}{dx}(x\,y)=\frac{d}{dx}(e).$$

The derivative of $$e^{y}$$ is $$e^{y}\dfrac{dy}{dx}$$ (chain rule). The derivative of $$x\,y$$ is $$y+x\dfrac{dy}{dx}$$ (product rule). The derivative of the constant $$e$$ is $$0.$$ Hence

$$e^{y}\frac{dy}{dx}+y+x\frac{dy}{dx}=0.$$

Collecting the $$\dfrac{dy}{dx}$$ terms yields

$$(e^{y}+x)\frac{dy}{dx}+y=0.$$

Solving for $$\dfrac{dy}{dx}$$ gives

$$\frac{dy}{dx}=-\frac{y}{\,e^{y}+x\,}.$$

At $$x=0,\;y=1$$ we obtain

$$\left.\frac{dy}{dx}\right|_{(0,1)}=-\frac{1}{e^{1}+0}=-\frac{1}{e}.$$

Thus the first component of the ordered pair is $$-\dfrac1e.$$

Next we find the second derivative $$\dfrac{d^{2}y}{dx^{2}}.$$ We start from the differentiated equation

$$e^{y}\frac{dy}{dx}+y+x\frac{dy}{dx}=0$$

and differentiate it once more with respect to $$x$$. Each term is handled carefully:

• The derivative of $$e^{y}\dfrac{dy}{dx}$$ is $$e^{y}\left(\frac{dy}{dx}\right)^{2}+e^{y}\frac{d^{2}y}{dx^{2}}.$$ The first part comes from differentiating $$e^{y}$$ (chain rule), the second part from differentiating $$\dfrac{dy}{dx}.$$
• The derivative of $$y$$ is $$\dfrac{dy}{dx}.$$
• The derivative of $$x\dfrac{dy}{dx}$$ is $$\dfrac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}$$ (product rule).
• The derivative of the right‐hand side $$0$$ is $$0.$$

Adding these pieces we get

$$e^{y}\left(\frac{dy}{dx}\right)^{2}+e^{y}\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}+\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}=0.$$

That is

$$e^{y}\left(\frac{dy}{dx}\right)^{2}+e^{y}\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}=0.$$

Factor the second‐derivative terms:

$$(e^{y}+x)\frac{d^{2}y}{dx^{2}}+e^{y}\left(\frac{dy}{dx}\right)^{2}+2\frac{dy}{dx}=0.$$

Now solve for $$\dfrac{d^{2}y}{dx^{2}}:$$

$$\frac{d^{2}y}{dx^{2}}=-\frac{e^{y}\left(\dfrac{dy}{dx}\right)^{2}+2\dfrac{dy}{dx}}{e^{y}+x}.$$

We evaluate this at $$x=0,\;y=1.$$ We already know $$\dfrac{dy}{dx}=-\dfrac1e.$$ Hence

$$e^{y}=e,\qquad\left(\frac{dy}{dx}\right)^{2}=\left(-\frac1e\right)^{2}=\frac1{e^{2}}.$$

Compute the numerator:

$$e^{y}\left(\frac{dy}{dx}\right)^{2}+2\frac{dy}{dx}=e\left(\frac1{e^{2}}\right)+2\left(-\frac1e\right)=\frac1e-\frac2e=-\frac1e.$$

The denominator is

$$e^{y}+x=e+0=e.$$

Therefore

$$\left.\frac{d^{2}y}{dx^{2}}\right|_{(0,1)}=-\frac{-\dfrac1e}{e}=\frac1{e^{2}}.$$

Thus the ordered pair is

$$\left(\frac{dy}{dx},\frac{d^{2}y}{dx^{2}}\right)=\left(-\frac1e,\;\frac1{e^{2}}\right).$$

Hence, the correct answer is Option B.