= Solution
The inverse on this <Weierstrass equation of an elliptic curve> is $(x,y)\mapsto(x,-y-1)$. A line $y=\lambda x+\nu$ through the points being added has third intersection $R$ with $x(R)=\lambda^2-x_1-x_2$, and the sum is $-R$.
At $P=(0,0)$, implicit differentiation gives $\lambda=(3x^2-1)/(2y+1)=-1$. The tangent is $y=-x$. Substitution into the equation gives $x^2=x^3$, so its third intersection is $(1,-1)$, whose inverse is $(1,0)$. Next the line through $(0,0)$ and $(1,0)$ is $y=0$; its remaining intersection is $(-1,0)$, whose inverse is $(-1,-1)$. Consequently
$$
\boxed{2P=(1,0),\qquad3P=(-1,-1).}
$$
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