Solve partialu/partialy=frac{-x*((y^{2}+z^2)^2)}{y^2+z^2}=

$$\frac { \partial u } { \partial y } = \frac { - x \cdot ( ( y ^ { 2 } + z ^ { 2 } ) ^ { 2 } ) } { y ^ { 2 } + z ^ { 2 } } =$$

Solve for x

$$x=0$$

$$z\neq 0\text{ or }y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y^{2}+z^{2}.

$$\left(y^{2}+z^{2}\right)\frac{\mathrm{d}}{\mathrm{d}y}(u)=\left(-x\right)\left(y^{2}+z^{2}\right)^{2}$$

Use the distributive property to multiply y^{2}+z^{2} by \frac{\mathrm{d}}{\mathrm{d}y}(u).

$$y^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)+z^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)=\left(-x\right)\left(y^{2}+z^{2}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y^{2}+z^{2}\right)^{2}.

$$y^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)+z^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)=\left(-x\right)\left(\left(y^{2}\right)^{2}+2y^{2}z^{2}+\left(z^{2}\right)^{2}\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$y^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)+z^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)=\left(-x\right)\left(y^{4}+2y^{2}z^{2}+\left(z^{2}\right)^{2}\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$y^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)+z^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)=\left(-x\right)\left(y^{4}+2y^{2}z^{2}+z^{4}\right)$$

Use the distributive property to multiply -x by y^{4}+2y^{2}z^{2}+z^{4}.

$$y^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)+z^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)=\left(-x\right)y^{4}+2\left(-x\right)y^{2}z^{2}+\left(-x\right)z^{4}$$

Swap sides so that all variable terms are on the left hand side.

$$\left(-x\right)y^{4}+2\left(-x\right)y^{2}z^{2}+\left(-x\right)z^{4}=y^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)+z^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)$$

Multiply 2 and -1 to get -2.

$$-xy^{4}-2xy^{2}z^{2}-xz^{4}=y^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)+z^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)$$

Combine all terms containing x.

$$\left(-y^{4}-2y^{2}z^{2}-z^{4}\right)x=y^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)+z^{2}\frac{\mathrm{d}}{\mathrm{d}y}(u)$$

The equation is in standard form.

$$\left(-y^{4}-2y^{2}z^{2}-z^{4}\right)x=0$$

Divide 0 by -y^{4}-2y^{2}z^{2}-z^{4}.

$$x=0$$

Solve for u

$$u\in \mathrm{R}$$

$$\left(y\neq 0\text{ or }z\neq 0\right)\text{ and }x=0$$