For the function $f_\lambda(x)=4\lambda x^3-36\lambda x^2+36x+48$ to be increasing for all $x\in\mathbb R$, we need $f_\lambda'(x)\ge 0 \quad \text{for all } x\in\mathbb R.$

Differentiate: $f_\lambda'(x)=12\lambda x^2-72\lambda x+36.$ Factor out $12$: $f_\lambda'(x)=12\big(\lambda x^2-6\lambda x+3\big).$

So we require $\lambda x^2-6\lambda x+3\ge 0 \quad \forall x\in\mathbb R.$

This is a quadratic in $x$. For it to be non-negative for all real $x$, two conditions are needed:

• coefficient of $x^2$ must be positive or zero,
• discriminant must be $\le 0$.

Here coefficient of $x^2$ is $\lambda$, so we need $\lambda\ge 0.$

Discriminant: $\Delta = (-6\lambda)^2-4(\lambda)(3)=36\lambda^2-12\lambda=12\lambda(3\lambda-1).$ For non-negativity for all $x$, $\Delta\le 0.$ Since $\lambda\ge 0$, this gives

Therefore, the largest such value is $\lambda^*=\frac13.$

Now compute $f_{\lambda^*}(x)$:

Evaluate at $x=1$: f_{1/3}(1)=\frac43-12+36+48=rac43+72=rac{220}{3}.

Evaluate at $x=-1$: $f_{1/3}(-1)=-\frac43-12-36+48=-\frac43+0=-\frac43.$

Add them: f_{1/3}(1)+f_{1/3}(-1)=\frac{220}{3}-\frac{4}{3}=rac{216}{3}=72.

Hence the correct option is $\boxed{72}.$