$\begin{array}{lll}& & \text{Implicit Differentiation Steps}\\ & & x^2+y^2=9\\ \text{•}& & \text{Rewrite}y\text{as a function}y\left(x\right)\text{:}\\ & & x^2+{y\left(x\right)}^2=9\\ \text{•}& & \text{Differentiate the left side}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2+{y\left(x\right)}^2\right)\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(f\left(x\right)+g\left(x\right)\right)=\frac{\ⅆ}{\ⅆx}f\left(x\right)+\frac{\ⅆ}{\ⅆx}g\left(x\right)\\ & & f\left(x\right)=x^2\\ & & g\left(x\right)={y\left(x\right)}^2\\ & & \text{This gives:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)+\frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^2\right)\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}\left(x^2\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\partial }{\partial x}\left(x^n\right)=n{x}^{n-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)=2\cdot x^1\\ & \text{◦}& \text{So,}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)=2\cdot x\\ & & \text{We can rewrite the derivative as:}\\ & & \left(2x\right)+\frac{\ⅆ}{\ⅆx}\left({y\left(x\right)}^2\right)\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{chain}}\text{rule to the term}{y\left(x\right)}^2\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{chain}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)=\mathrm{f\text{'}}\left(g\left(x\right)\right)\left(\frac{\ⅆ}{\ⅆx}g\left(x\right)\right)\\ & \text{◦}& \text{Outside function}\\ & & f\left(v\right)=v^2\\ & \text{◦}& \text{Inside function}\\ & & g\left(x\right)=y\left(x\right)\\ & \text{◦}& \text{Derivative of outside function}\\ & & \frac{\ⅆ}{\ⅆv}f\left(v\right)=2v\\ & \text{◦}& \text{Apply composition}\\ & & \mathrm{f\text{'}}\left(g\left(x\right)\right)=2y\left(x\right)\\ & \text{◦}& \text{Derivative of inside function}\\ & & \frac{\ⅆ}{\ⅆx}g\left(x\right)=\frac{\ⅆ}{\ⅆx}y\left(x\right)\\ & \text{◦}& \text{Put it all together}\\ & & \left(\frac{\ⅆ}{\ⅆx}f\left(g\left(x\right)\right)\right)\left(\frac{\ⅆ}{\ⅆx}g\left(x\right)\right)=2y\left(x\right)\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\\ & & \text{This gives:}\\ & & 2x+\left(2y\left(x\right)\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\right)\\ \text{•}& & \text{The final result is}\\ & & 2x+2\cdot y\left(x\right)\cdot \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)\\ \text{•}& & \text{Differentiate the right side}\\ & & \frac{\ⅆ}{\ⅆx}9\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{constant}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}9\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \frac{\ⅆC}{\ⅆx}=0\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}9=0\\ & & \text{We can now rewrite the derivative as:}\\ & & 0\\ \text{•}& & \text{Rewrite}\frac{\ⅆ}{\ⅆx}y\left(x\right)\text{as}\mathrm{y\text{'}}\text{and solve for}\mathrm{y\text{'}}\\ & & 2x+2\cdot y\cdot \mathrm{y\text{'}}=0\\ \text{•}& & \text{Subtract}2\cdot x\text{from both sides}\\ & & 2\cdot x+2\cdot y\cdot \mathrm{y\text{'}}-2\cdot x=0-2\cdot x\\ \text{•}& & \text{Simplify}\\ & & 2\cdot y\cdot \mathrm{y\text{'}}=\left(\mathrm{-2}\right)\cdot x\\ \text{•}& & \text{Divide both sides by}2\cdot y\\ & & \frac{\mathrm{y\text{'}}\cdot \left(2\cdot y\right)}{2\cdot y}=\frac{\left(\mathrm{-2}\right)\cdot x}{2\cdot y}\\ \text{•}& & \text{Simplify}\\ & & \mathrm{y\text{'}}=\frac{-2x}{2\cdot y}\\ \text{•}& & \text{Reduce fraction by gcd}\\ & & \mathrm{y\text{'}}=-\frac{x}{y}\end{array}$

$\begin{array}{lll}& & 3\cdot x^2+2\cdot x+1\\ \text{•}& & \text{Add and subtract}\frac{1}{3}\cdot {\left(\frac{2}{2}\right)}^2\\ & & 3x^2+2x+\frac{1}{3}\cdot {\left(\frac{2}{2}\right)}^2-\frac{1}{3}\cdot {\left(\frac{2}{2}\right)}^2+1\\ \text{•}& & \text{Simplify terms}\\ & & 3x^2+2x+\frac{1}{3}-\frac{1}{3}+1\\ \text{•}& & \text{The first}3\text{terms can be regrouped as a perfect square}\\ & & 3{\left(x+\frac{1}{3}\right)}^2-\frac{1}{3}+1\\ \text{•}& & \text{Simplify the remaining term}\\ & & 3{\left(x+\frac{1}{3}\right)}^2+\frac{2}{3}\end{array}$

$\begin{array}{cccccccccc}& & & \underset{\—}{ }& \underset{\—}{ }& \underset{\—}{ }& \underset{\—}{\.}& \underset{\—}{6}& \underset{\—}{6}& \underset{\—}{6}\\ & 3& & \)& & 2& \.& 0& 0& 0\\ & & & & & & \underset{\—}{1}& \underset{\—}{8}\\ & & & & & & & 2& 0\\ & & & & & & & \underset{\—}{1}& \underset{\—}{8}\\ & & & & & & & & 2& 0\\ & & & & & & & & \underset{\—}{1}& \underset{\—}{8}\\ & & & & & & & & & 2\\ \=0.666+\frac{1}{1500}\end{array}$

$\begin{array}{cc}\stackrel{\phantom{z^2}}{x+3}& \begin{array}{cccc}& \phantom{\mathrm{PP}}3x& \phantom{\mathrm{PP}}\mathrm{-7}& \\ \)\phantom{x^2}& \phantom{1}3x^2& \phantom{1}\+2x& \phantom{1}\+1\\ & \multicolumn{2}{c}{\frac{3x^2+9x}{\phantom{\.}}}& \\ & & \multicolumn{2}{c}{-7x+1}& & \\ & & \multicolumn{2}{c}{\frac{-7x-21}{\phantom{\.}}}& & \\ & & & 22\hfill & & & \end{array}\\ \=3x-7+\frac{22}{x+3}\end{array}$