Let $f(x)= (1+ \frac{|x|^2}{N(N-2)})^{\frac{N-2}{2}}, (x\in \mathbb R^N).$
My Question is: How to show that $\Delta f + |f|^{4/(N-2)}f=0$?
My attempt: Suppose take $N=3.$ Note that $\frac{\partial f (x)}{\partial x_1}= -\frac{1}{N}(1+ \frac{|x|^2}{N(N-2)})^{\frac{-N}{2}} x_1$ $x=(x_1, x_2, x_3)\in \mathbb R^3.$ So $$\frac{\partial^2 f (x)}{\partial x_1^2}= -\frac{1}{N}(1+ \frac{|x|^2}{N(N-2)})^{\frac{-N}{2}} + (1+ \frac{|x|^2}{N(N-2)})^{\frac{-N-2}{2}} \frac{1}{N(N-2)} x_1^2$$ Hence, $$\Delta f (x) = -\frac{3}{N}(1+ \frac{|x|^2}{N(N-2)})^{\frac{-N}{2}} + (1+ \frac{|x|^2}{N(N-2)})^{\frac{-N-2}{2}} \frac{1}{N(N-2)} |x|^2$$