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Explanation:

Step1: Differentiate the given equation implicitly

Differentiate $x^{2}+xy + y^{2}=7$ with respect to $x$. Using the sum - rule of differentiation, we have $\frac{d}{dx}(x^{2})+\frac{d}{dx}(xy)+\frac{d}{dx}(y^{2})=\frac{d}{dx}(7)$. The derivative of $x^{2}$ with respect to $x$ is $2x$. For $xy$, using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x$ and $v = y$, we get $y + x y^\prime$. For $y^{2}$, using the chain - rule, we get $2y y^\prime$. And the derivative of a constant $7$ is $0$. So, $2x + y+xy^\prime+2y y^\prime = 0$.

Step2: Solve for $y^\prime$

Group the terms with $y^\prime$ together: $xy^\prime+2y y^\prime=-2x - y$. Factor out $y^\prime$: $y^\prime(x + 2y)=-(2x + y)$. Then $y^\prime=\frac{-(2x + y)}{x + 2y}=\frac{-2x - y}{x + 2y}$.

Step3: Differentiate $y^\prime$ to find $y^{\prime\prime}$

Using the quotient - rule $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u=-2x - y$ and $v=x + 2y$. First, find $u^\prime=-2 - y^\prime$ and $v^\prime=1 + 2y^\prime$. Then $y^{\prime\prime}=\frac{(-2 - y^\prime)(x + 2y)-(-2x - y)(1 + 2y^\prime)}{(x + 2y)^{2}}$. Substitute $y^\prime=\frac{-2x - y}{x + 2y}$ into the above formula: [ \begin{align*} y^{\prime\prime}&=\frac{(-2-\frac{-2x - y}{x + 2y})(x + 2y)-(-2x - y)(1 + 2\frac{-2x - y}{x + 2y})}{(x + 2y)^{2}}\ &=\frac{(-2(x + 2y)+2x + y)(x + 2y)-(-2x - y)(x + 2y-4x - 2y)}{(x + 2y)^{3}}\ &=\frac{(-2x-4y + 2x + y)(x + 2y)-(-2x - y)(-3x)}{(x + 2y)^{3}}\ &=\frac{(-3y)(x + 2y)-3x(2x + y)}{(x + 2y)^{3}}\ &=\frac{-3xy-6y^{2}-6x^{2}-3xy}{(x + 2y)^{3}}\ &=\frac{-6x^{2}-6xy - 6y^{2}}{(x + 2y)^{3}}\ &=\frac{-6(x^{2}+xy + y^{2})}{(x + 2y)^{3}} \end{align*} ] Since $x^{2}+xy + y^{2}=7$, then $y^{\prime\prime}=\frac{-42}{(x + 2y)^{3}}$.

Answer:

$y^\prime=\frac{-2x - y}{x + 2y}$, $y^{\prime\prime}=\frac{-42}{(x + 2y)^{3}}$