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Equation (13) expresses the metric on an embedded hypersurface given by the relations $y^k = y^k(x^a)$. However, the equation for the inverse metric (4-th equation) is in general not correct. Take for example a hypersurface defined by: $y^1 = x^1$, $y^2 = x^2$, $y^3 = x^2$. In our case, the partial derivative of $x^2$ with respect to $y^2$ or $y^3$ ...
Let there be given a manifold $(M,\nabla)$ equipped with a (not necessarily torsionfree) tangent bundle connection $\nabla$. OP is asking: Is it possible that the local coordinate expression for the covariant derivative of a co-vector/one-form $\lambda =\lambda_a \mathrm{d}x^a$ (in some local coordinate system $x^a$) could be on the form $$\tag{1} ... 4 What you said is only true if the hypersurface is space-like or time-like. If a non-null hypersurface is defined by f(x) =  constant, then the normal to the hypersurface is given by$$ n_\alpha \propto \partial_\alpha f $$The fact that the hypersurface is non-null implies$$ g^{\alpha\beta} \partial_\alpha f \partial_\beta f = \varepsilon\neq 0 $$... 0 The Einstein equations are some of the most complicated PDE's people study. There is no shortcut for this, you just have to do all the horrific algebra. Start with the trace-reversed Einstein equation $$R_{\mu \nu}=8\pi G(T_{\mu \nu}-\frac{1}{2}Tg_{\mu \nu})$$ Use the equation for Ricci in terms of the Christoffel Connection ... 0 For a start, you might have a look at the paper "Differential forms for scientists" by J.B. Perot, Journal of Computational Physics, 2014, Vol. 257 Part B, I found it useful. 4 I'll write this as an answer so that the math is more clear. So given an (p,q)-tensor T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}, this one transforms as:$$T'^{\mu'_1\cdots\mu'_p}{}_{\nu'_1\cdots\nu'_q}=\frac{\partial x^{\mu'_1}}{\partial x^{\mu_1}}\cdots \frac{\partial x^{\mu'_p}}{\partial x^{\mu_p}}\frac{\partial x^{\nu_1}}{\partial ...