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For the following functions find $$\frac{d y}{d x}$$

$$x^{3}+y^{3}=1$$
$$\mathbf{e}^{x / y}=x y$$
$$\frac{1}{x}+\frac{1}{y}=2$$

$$x^{3}+y^{3}=1 \rightarrow 3 x^{2}+3 y^{2} \frac{d y}{d x}=0 \rightarrow x^{2}+y^{2} \frac{d y}{d x}=0$$

$$\frac{d y}{d x}=\frac{-x^{2}}{y^{2}}$$

$$e^{x / y}=x-y \longrightarrow e^{x / y}\left(\frac{\left(1)(y)-\frac{d y}{d x}(x)\right.}{y^{2}}\right)=1-\frac{d y}{d x}$$

$$e^{x / y}\left(y-x \frac{d y}{d x}\right)=y^{2}-y^{2} \frac{d y}{d x}$$

$$\rightarrow y e^{x / y}-x e^{x / y} \frac{d y}{d x}=y^{2}-y^{2} \frac{d y}{d x}$$

$$-x e^{x / y} \frac{d y}{d x}+y^{2} \frac{d y}{d x}=y^{2}-y e^{x / y}$$

$$\frac{d y}{d x}=\frac{y^{2}-y e^{x / y}}{-x e^{x / y}+y^{2}}$$

$$\frac{1}{x}+\frac{1}{y}=2 \longrightarrow \frac{-1}{x^{2}}+\frac{-\frac{d y}{d x}}{y^{2}}=0$$

$$\frac{\frac{-d y}{d x}}{y^{2}}=1 / x^{2} \rightarrow$$

$$\frac{-d y}{d x}=\frac{y^{2}}{x^{2}} \rightarrow \frac{d x}{d x}=\frac{-y^{2}}{x^{2}}$$

find $$\frac{d y}{d x}$$ by implicit differentiation:

$$y \cos x=x^{2}+y^{2}$$

$$y \cos x=x^{2}+y^{2} \rightarrow \frac{d y}{d x}(\cos x)+(-\sin x)(y)=2 x+2 y \frac{d y}{d x}$$

$$\frac{d y}{d x} \cos x-2 y \frac{d y}{d x}=2 x+y \sin x$$

$$\frac{d y}{d x}=\frac{2 x+y \sin x}{\cos x-2 y}$$

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