= Solution
An <inflection point of an algebraic plane curve> is a smooth point where the tangent has <intersection multiplicity> at least three. For a homogeneous defining polynomial $P$ of degree $d$, its <Hessian curve> is defined by the determinant of the matrix of second partial derivatives, of degree $3(d-2)$. In characteristic zero, the smooth inflection points are exactly the smooth points of $C\cap\operatorname{Hess}(C)$.
For a local verification, move a smooth point to $[0:0:1]$ and its tangent to $y=0$, and scale so $P(x,y,1)=y+ax^2+bxy+cy^2+\cdots$. The tangent contact is at least three exactly when $a=0$. The homogeneous Euler identities give the Hessian matrix at that point as $\begin{pmatrix}2a&b&0\\b&2c&d-1\\0&d-1&0\end{pmatrix}$, whose determinant is $-2a(d-1)^2$. This proves the criterion for $d>1$.
For the given cubic, the singularity equations are $x^2+\lambda yz=y^2+\lambda xz=z^2+\lambda xy=0$. No coordinate can vanish at a projective solution, since that would force all three to vanish. Multiplying the equations therefore gives $1=(-\lambda)^3$, so singularity requires $\lambda^3=-1$. Conversely, under that condition the point $[1:1:-1/\lambda]$ satisfies all three equations, and the Euler identity places it on the cubic. Thus
$$
\boxed{C\text{ is nonsingular}\iff\lambda^3\ne-1.}
$$
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