If $\displaystyle \left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}=\mathrm{x} \mathrm{y}$, then find $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}$.

$$\begin{gathered} \left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}=\mathrm{xy} \text { gives } \\ 2\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\left[\mathbf{2} \boldsymbol{\mathrm{x}}+\mathbf{2} \boldsymbol{\mathrm{y}} \frac{\boldsymbol{d} \boldsymbol{\mathrm{y}}}{\boldsymbol{d} \boldsymbol{\mathrm{x}}}\right]=\boldsymbol{\mathrm{x}} \frac{\boldsymbol{d} \boldsymbol{\mathrm{y}}}{\boldsymbol{d} \boldsymbol{\mathrm{x}}}+\boldsymbol{\mathrm{y}} \\ \Rightarrow\left[4 \mathrm{y}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)-\mathrm{x}\right] \frac{\boldsymbol{d} \boldsymbol{\mathrm{y}}}{\boldsymbol{d} \boldsymbol{\mathrm{x}}}=\mathrm{y}-4 \mathrm{x}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) \\ \Rightarrow \frac{\boldsymbol{d} \boldsymbol{\mathrm{y}}}{\boldsymbol{d} \boldsymbol{\mathrm{x}}}=\frac{\boldsymbol{\mathrm{y}}-\mathbf{4} \boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}^{2}+\boldsymbol{\mathrm{y}}^{2}\right)}{\mathbf{4} \boldsymbol{\mathrm{y}}\left(\boldsymbol{\mathrm{x}}^{2}+\boldsymbol{\mathrm{y}}^{2}\right)-\boldsymbol{\mathrm{x}}} \end{gathered} $$