1. Let $z=x+iy,$ where $x,y\in\mathbb{R}$ and $z\neq 0$.

2. Compute both sides of the given equation: $z^2=i|z|^2.$

   We have $z^2=(x+iy)^2=x^2-y^2+2ixy,$ and $|z|^2=x^2+y^2.$ So, $i|z|^2=i(x^2+y^2).$

3. Equate real and imaginary parts: $x^2-y^2+2ixy=i(x^2+y^2).$

   Comparing real parts: $x^2-y^2=0$ $\Rightarrow x^2=y^2$ $\Rightarrow y=\pm x.$

   Comparing imaginary parts: $2xy=x^2+y^2.$

4. Now test the two cases from $y=\pm x$.

   Case 1: $y=x$ $2x\cdot x=x^2+x^2$ $2x^2=2x^2,$ which is true.

   Case 2: $y=-x$ $2x(-x)=x^2+x^2$ $-2x^2=2x^2$ $\Rightarrow 4x^2=0$ $\Rightarrow x=0.$ Then $y=0$, so $z=0$, which is not allowed since $z\neq 0$.

5. Hence the only possible case is $y=x.$ Therefore, $z$ lies on the line $y=x.$

6. Check options:

   • A: $y=-x$ ❌
   • B: real axis ❌
   • C: $y=x$ ✅
   • D: imaginary axis ❌

Therefore, the correct answer is C.