1. Given differential equation

We have

with initial condition $y(1)=-1.$

We need to find $\phi\left(\frac{y^2}{4}\right).$

2. Substitute a standard homogeneous variable

Let $v=\frac{y^2}{x^2}.$ Then $y^2=vx^2.$ Differentiating both sides w.r.t. $x$: $2y\frac{dy}{dx}=x^2\frac{dv}{dx}+2vx.$ So $y\frac{dy}{dx}=\frac{x^2}{2}\frac{dv}{dx}+vx.$

Now substitute into the given equation: $\frac{x^2}{2}\frac{dv}{dx}+vx=x\left[v+\frac{\phi(v)}{\phi'(v)}\right].$ Cancel $xv$ from both sides: $\frac{x^2}{2}\frac{dv}{dx}=x\frac{\phi(v)}{\phi'(v)}.$ Since $x>0$, divide by $x$: $\frac{x}{2}\frac{dv}{dx}=\frac{\phi(v)}{\phi'(v)}.$ Thus, $\frac{dv}{dx}=\frac{2}{x}\frac{\phi(v)}{\phi'(v)}.$

3. Separate variables

Rearrange: $\frac{\phi'(v)}{\phi(v)}\,dv=2\frac{dx}{x}.$ Integrate both sides: $\int \frac{\phi'(v)}{\phi(v)}\,dv=\int 2\frac{dx}{x}.$ This gives $\ln \phi(v)=2\ln x + C.$ Hence, $\phi(v)=Cx^2.$

4. Use the initial condition

At $x=1$, $y=-1$, so $v=\frac{y^2}{x^2}=\frac{1}{1}=1.$ Therefore, $\phi(1)=C\cdot 1^2=C.$ So $\phi(v)=\phi(1)x^2.$ That is, $\phi\left(\frac{y^2}{x^2}\right)=\phi(1)x^2.$

5. Evaluate at the required expression

We need $\phi\left(\frac{y^2}{4}\right).$ Notice that $\frac{y^2}{4}=\frac{y^2}{x^2} \quad \text{when } x=2.$ So at $x=2$, $\phi\left(\frac{y^2}{4}\right)=\phi\left(\frac{y^2}{x^2}\right)=\phi(1)x^2=\phi(1)\cdot 2^2=4\phi(1).$

6. Match with options

Thus, $\boxed{\phi\left(\frac{y^2}{4}\right)=4\phi(1)}.$

So the correct option is: $\boxed{\text{B}}$