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Step 2 Next, find \( f_{y}(x, y) \). \[ \begin{array}{l} f_{y}(x, y)=\frac{\partial}{\partial y}\left(\frac{x}{x^{2}+y^{2}}\right) \\ =\frac{\left(\left(x^{2}+y^{2}\right) \cdot 0-20-2 x y\right)}{\left(x^{2}+y^{2}\right)^{2}} \\ =\frac{-2 \cdot x y}{\left(x^{2}+y^{2}\right)^{2}} \\ \end{array} \] Step 3 Therefore, \[ \begin{aligned} \nabla f(x, y) & =f_{x}(x, y) \hat{i}+f_{y}(x, y) j \\ & =\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}} \mathbf{i}+\frac{-2 x y}{\left(x^{2}+y^{2}\right)^{2}} \mathbf{j} . \end{aligned} \] The level curve for \( c=\frac{1}{4} \) is given by \( f(x, y)= \)

          Step 2
Next, find \( f_{y}(x, y) \).
\[
\begin{array}{l}
f_{y}(x, y)=\frac{\partial}{\partial y}\left(\frac{x}{x^{2}+y^{2}}\right) \\
=\frac{\left(\left(x^{2}+y^{2}\right) \cdot 0-20-2 x y\right)}{\left(x^{2}+y^{2}\right)^{2}} \\
=\frac{-2 \cdot x y}{\left(x^{2}+y^{2}\right)^{2}} \\
\end{array}
\]

Step 3
Therefore,
\[
\begin{aligned}
\nabla f(x, y) & =f_{x}(x, y) \hat{i}+f_{y}(x, y) j \\
& =\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}} \mathbf{i}+\frac{-2 x y}{\left(x^{2}+y^{2}\right)^{2}} \mathbf{j} .
\end{aligned}
\]

The level curve for \( c=\frac{1}{4} \) is given by \( f(x, y)= \)

Step 2
Next, find fy(x, y).

fy(x, y)=(∂)/(∂ y)((x)/(x^2+y^2))

=(((x^2+y^2) · 0-20-2 x y))/((x^2+y^2)^2)

=(-2 · x y)/((x^2+y^2)^2)

Step 3
Therefore,

∇ f(x, y) =fx(x, y) î+fy(x, y) j
=(y^2-x^2)/((x^2+y^2)^2)𝐢+(-2 x y)/((x^2+y^2)^2)𝐣 .

The level curve for c=(1)/(4) is given by f(x, y)=

Added by Nicholas C.

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