1. The problem is to understand and interpret the expression $$\frac{\partial y^n}{\partial y} = y \delta u$$. 2. Here, $$\frac{\partial y^n}{\partial y}$$ denotes the partial derivative of $$y^n$$ with respect to $$y$$. 3. The formula for the derivative of $$y^n$$ with respect to $$y$$ is $$\frac{d}{dy} y^n = n y^{n-1}$$. 4. The right side of the equation is $$y \delta u$$, where $$\delta u$$ likely represents a small change or variation in $$u$$. 5. To verify or interpret this, we note that $$\frac{\partial y^n}{\partial y} = n y^{n-1}$$, so the equation implies $$n y^{n-1} = y \delta u$$. 6. This can be rearranged to $$\delta u = \frac{n y^{n-1}}{y} = n y^{n-2}$$. 7. Therefore, the expression relates the partial derivative of $$y^n$$ to a product involving $$y$$ and a variation $$\delta u$$, where $$\delta u = n y^{n-2}$$. Final answer: $$\frac{\partial y^n}{\partial y} = n y^{n-1}$$ and $$\delta u = n y^{n-2}$$.