Show that if $v(t,\|x\|)$ solves the heat equation on the set $Q \subset (0,+\infty)\times\mathbb{R}^n$, then the function $v(\lambda^2t,\lambda\|x\|)$ solves the heat equation on the set $Q_\lambda=\{(t,x)\in (0,+\infty)\times\mathbb{R}^n:(\lambda^2t,\lambda\|x\|)\in Q\}.$
My attempt: Let $w(t,x)=v(\lambda^2t,\lambda\|x\|)=v(f_1(t),f_2(x)).$ Then by applying the chain rule to obtain the partial derivatives we get $$w_t(t,x)=\frac{\partial v}{\partial f_1}\frac{\partial f_1}{\partial x}=\lambda^2\frac{\partial v}{\partial f_1}$$ $$w_{x_i}=\frac{\partial v}{\partial f_2}\frac{\partial f_2}{\partial x}=\lambda\frac{\partial v}{\partial f_2}x_i(x_1^2+...+x_n^2)^{-\frac{1}{2}}$$ $$w_{x_ix_i}=\lambda(x_1^2+...+x_n^2)^{-\frac{1}{2}}\frac{\partial v}{\partial f_2}-\lambda x_i^2(x_1^2+...+x_n^2)^{-\frac{3}{2}}\frac{\partial v}{\partial f_2}+\lambda^2x_i^2(x_1^2+...+x_n^2)\frac{\partial^2 v}{\partial^2 f_2}$$
which gives us $$\triangle_xw=\lambda \frac{1}{\|x\|}\frac{\partial v}{\partial f_2}-\lambda \frac{1/}{\|x\|^2}\frac{\partial v}{\partial f_2}+\|x\|^4\lambda^2\frac{\partial^2 v}{\partial^2 f_2}$$
I'm not sure this is the right approach since plugging the above into the heat equation doesnt satisfy it.